Electrostatics MCQs with Answers – Part 2 (NEET Physics)

Explanation: \(\textbf{Infer the source-charge sign:}\) The field at \(x=-a\) points towards positive \(x\). This is the direction from the observation point towards the origin. A negative point charge produces an electric field directed towards itself. Therefore, the source charge is negative. A positive source at the origin would instead produce a field towards negative \(x\) at \(x=-a\). \(\textbf{Determine the field direction at }x=+2a\textbf{:}\) The point \(x=+2a\) lies to the right of the negative source. The electric field points towards the source, which is towards negative \(x\). The given magnitude is \(200\,\text{N C}^{-1}\). Therefore, \[ \vec E=-200\hat i\,\text{N C}^{-1} \] \(\textbf{Verify the inverse-square magnitude:}\) The first observation point is at distance \(a\). The second is at distance \(2a\). Coulomb's point-charge field relation gives \[ \frac{E_2}{E_1}=\left(\frac{a}{2a}\right)^2 \] \[ \frac{E_2}{E_1}=\frac14 \] Hence, \[ E_2=\frac{800}{4}=200\,\text{N C}^{-1} \] This agrees with the stated second measurement. \(\textbf{Calculate the force on the negative test charge:}\) The test charge is \[ q_t=-3\times10^{-6}\,\text{C} \] The electric force is \[ \vec F=q_t\vec E \] Substituting, \[ \vec F=(-3\times10^{-6})(-200\hat i) \] \[ \vec F=+600\times10^{-6}\hat i\,\text{N} \] \[ \vec F=+0.60\hat i\,\text{mN} \] \(\textbf{Physical check:}\) Both the source and test charge are negative. They repel each other. Since the test charge is to the right of the source, the repulsive force must point towards positive \(x\). This agrees with the calculation. \(\textbf{Final answer:}\) \[ \boxed{Q\lt0,\qquad \vec F=+0.60\hat i\,\text{mN}} \]

Explanation: \(\textbf{Locate the source relative to the observation point:}\) The source charge is at \[ (a,2a) \] The observation point is the origin. The vector from the origin towards the source is \[ \vec r_s=a\hat i+2a\hat j \] Its magnitude is \[ r=\sqrt{a^2+(2a)^2} \] \[ r=\sqrt5a \] \(\textbf{Determine the field direction:}\) The source charge is negative. Therefore, its electric field at the origin points towards the source, along \(a\hat i+2a\hat j\). The unit vector in this direction is \[ \hat u=\frac{\hat i+2\hat j}{\sqrt5} \] \(\textbf{Calculate the electric-field magnitude:}\) For a point charge, \[ E=k\frac{|Q|}{r^2} \] Substituting \(r=\sqrt5a\), \[ E=k\frac{Q}{(\sqrt5a)^2} \] \[ E=\frac{kQ}{5a^2} \] Using \(E_0=kQ/a^2\), \[ E=\frac{E_0}{5} \] \(\textbf{Construct the electric-field vector:}\) \[ \vec E=E\hat u \] \[ \vec E=\frac{E_0}{5}\frac{\hat i+2\hat j}{\sqrt5} \] \[ \vec E=\frac{E_0}{5\sqrt5}(\hat i+2\hat j) \] \(\textbf{Determine the force on the negative test charge:}\) The test charge is \(-q\). Therefore, \[ \vec F=(-q)\vec E \] Substituting, \[ \vec F=-q\frac{E_0}{5\sqrt5}(\hat i+2\hat j) \] \[ \vec F=-\frac{qE_0}{5\sqrt5}(\hat i+2\hat j) \] \(\textbf{Direction check:}\) The electric field points towards the negative source. However, the negative test charge is repelled by that source. Its force therefore points away from \((a,2a)\), opposite to the electric-field direction. \(\textbf{Final answer:}\) \[ \boxed{\vec E=\frac{E_0}{5\sqrt5}(\hat i+2\hat j),\qquad \vec F=-\frac{qE_0}{5\sqrt5}(\hat i+2\hat j)} \]

Explanation: \(\textbf{Original electric field:}\) For a point source, \[ E=k\frac{|Q|}{r^2} \] The original source is \(+Q\), so \[ E_i=\frac{kQ}{r^2} \] The original test charge has magnitude \(q\). Its force magnitude is \[ F_i=qE_i \] \[ F_i=\frac{kqQ}{r^2} \] \(\textbf{Modified source and observation distance:}\) The source charge increases to \(+3Q\). The observation distance becomes \(2r\). The new electric-field magnitude is \[ E_f=k\frac{3Q}{(2r)^2} \] \[ E_f=\frac{3kQ}{4r^2} \] Comparing with the initial field, \[ \frac{E_f}{E_i}=\frac{3kQ/(4r^2)}{kQ/r^2} \] \[ \frac{E_f}{E_i}=\frac34 \] \(\textbf{Effect of the changed test charge:}\) The replacement test charge has magnitude \(2q\). The new force magnitude is \[ F_f=(2q)E_f \] Substituting, \[ F_f=(2q)\frac{3kQ}{4r^2} \] \[ F_f=\frac{3kqQ}{2r^2} \] Therefore, \[ \frac{F_f}{F_i}=\frac{3kqQ/(2r^2)}{kqQ/r^2} \] \[ \frac{F_f}{F_i}=\frac32 \] \(\textbf{Conceptual distinction:}\) The electric field depends on the source charge and distance, but not on the test-charge magnitude. The force on the test charge depends on both the electric field and the test charge. Consequently, the field can decrease while the measured electric force increases. All charges are positive, so the force remains along the outward source-field direction. \(\textbf{Final answer:}\) \[ \boxed{\frac{E_f}{E_i}=\frac34,\qquad \frac{F_f}{F_i}=\frac32} \]

Explanation: \(\textbf{Choose a common field scale:}\) The distance from either source charge to \(P(0,a)\) is \[ r=\sqrt{a^2+a^2}=\sqrt2a \] Define \[ E_0=\frac{kQ}{2a^2} \] This is the field magnitude at \(P\) due to the source \(+Q\). \(\textbf{Field due to the left positive source:}\) The source \(+Q\) lies at \((-a,0)\). Its electric field at \(P\) points away from the source, upward and rightward. The corresponding unit vector is \[ \frac{\hat i+\hat j}{\sqrt2} \] Therefore, \[ \vec E_L=\frac{E_0}{\sqrt2}(\hat i+\hat j) \] \(\textbf{Initial field due to the right negative source:}\) The source \(-2Q\) lies at \((a,0)\). Its field at \(P\) points towards that source, downward and rightward. Its field magnitude is \(2E_0\). Thus, \[ \vec E_R=\frac{2E_0}{\sqrt2}(\hat i-\hat j) \] \(\textbf{Initial resultant:}\) \[ \vec E=\vec E_L+\vec E_R \] \[ \vec E=\frac{E_0}{\sqrt2}[(\hat i+\hat j)+2(\hat i-\hat j)] \] \[ \vec E=\frac{E_0}{\sqrt2}(3\hat i-\hat j) \] Its magnitude is \[ |\vec E|=\frac{E_0}{\sqrt2}\sqrt{3^2+(-1)^2} \] \[ |\vec E|=\sqrt5E_0 \] \(\textbf{After reversing the right source:}\) The right source becomes \(+2Q\). Its electric-field direction reverses, while its magnitude remains \(2E_0\). Therefore, \[ \vec E_R'=\frac{2E_0}{\sqrt2}(-\hat i+\hat j) \] The left-source contribution is unchanged. Hence, \[ \vec E'=\vec E_L+\vec E_R' \] \[ \vec E'=\frac{E_0}{\sqrt2}[(\hat i+\hat j)+2(-\hat i+\hat j)] \] \[ \vec E'=\frac{E_0}{\sqrt2}(-\hat i+3\hat j) \] The new magnitude is \[ |\vec E'|=\frac{E_0}{\sqrt2}\sqrt{(-1)^2+3^2} \] \[ |\vec E'|=\sqrt5E_0 \] The magnitudes are therefore equal. \(\textbf{Determine the angle between the fields:}\) Using the scalar product, \[ \cos\theta=\frac{\vec E\cdot\vec E'}{|\vec E||\vec E'|} \] \[ \cos\theta=\frac{(3)(-1)+(-1)(3)}{10} \] \[ \cos\theta=-\frac35 \] Since the cosine is negative, the angle is obtuse. \(\textbf{Final answer:}\) The resultant electric fields have equal magnitudes, but their directions make an obtuse angle.

Explanation: \(\textbf{Interpret the field data:}\) The measurements follow the inverse-square dependence for a point source: \[ E=\frac{kQ}{r^2} \] The field magnitude at \(2r\) is \[ E_1=\frac{E_0}{4} \] The field magnitude at \(3r\) is \[ E_2=\frac{E_0}{9} \] \(\textbf{Initial electric force:}\) The first test charge is negative and has magnitude \(q\). The magnitude of the force is \[ F_1=qE_1 \] Substituting, \[ F_1=q\frac{E_0}{4} \] \[ F_1=\frac{qE_0}{4} \] \(\textbf{Modified electric force:}\) At the second point, the test-charge magnitude is \(3q\). The electric-field magnitude there is \(E_0/9\). Therefore, \[ F_2=(3q)E_2 \] \[ F_2=3q\frac{E_0}{9} \] \[ F_2=\frac{qE_0}{3} \] \(\textbf{Compare force magnitudes:}\) \[ \frac{F_2}{F_1}=\frac{qE_0/3}{qE_0/4} \] \[ \frac{F_2}{F_1}=\frac43 \] \(\textbf{Determine force directions:}\) The source charge is positive, so its electric field points radially outward. The test charge is negative in both experiments. Its electric force therefore points radially inward, towards the positive source. Both observation points lie on the same radial line, so the force directions are identical. The force magnitude increases even though the electric-field magnitude decreases, because the test-charge magnitude increases by a larger factor. \(\textbf{Final answer:}\) \[ \boxed{\frac{F_2}{F_1}=\frac43\quad\text{(same radial direction)}} \]

Explanation: \(\textbf{Determine the point-charge field direction:}\) The source charge at the origin is negative. At positions on the positive \(x\)-axis, its electric field points towards the origin, along negative \(x\). The external uniform field points along positive \(x\). The two fields therefore oppose each other at \(x=a\). \(\textbf{Apply the zero-field condition:}\) The magnitude of the negative source's field at \(x=a\) is \[ E_Q(a)=\frac{kQ}{a^2} \] The total signed field there is \[ E_{\mathrm{net}}(a)=E_u-\frac{kQ}{a^2} \] The given condition is \[ E_{\mathrm{net}}(a)=0 \] Therefore, \[ E_u-\frac{kQ}{a^2}=0 \] \[ \frac{kQ}{a^2}=E_u \] \(\textbf{Field at the new position:}\) At \(x=2a\), the negative source produces a field of magnitude \[ E_Q(2a)=\frac{kQ}{(2a)^2} \] \[ E_Q(2a)=\frac{kQ}{4a^2} \] Using the cancellation condition, \[ E_Q(2a)=\frac{E_u}{4} \] This contribution points towards negative \(x\). The uniform field remains \(+E_u\hat i\). Hence, \[ \vec E_{\mathrm{net}}(2a)=\left(E_u-\frac{E_u}{4}\right)\hat i \] \[ \vec E_{\mathrm{net}}(2a)=+\frac34E_u\hat i \] \(\textbf{Force on the negative test charge:}\) For test charge \(-q\), \[ \vec F=(-q)\vec E_{\mathrm{net}} \] \[ \vec F=(-q)\left(\frac34E_u\hat i\right) \] \[ \vec F=-\frac34qE_u\hat i \] \(\textbf{Physical check:}\) At \(x=a\), the two field magnitudes are equal. At \(x=2a\), the point-charge field becomes one-fourth as strong, while the external uniform field remains unchanged. The net field therefore points along positive \(x\). The negative test charge experiences force in the opposite direction. \(\textbf{Final answer:}\) \[ \boxed{\vec E_{\mathrm{net}}=\frac34E_u\hat i,\qquad \vec F=-\frac34qE_u\hat i} \]

Explanation: \(\textbf{Examine the field directions between the charges:}\) The source charges have opposite signs. Between them, the positive source produces an electric field directed away from itself, towards positive \(x\). The negative source produces an electric field directed towards itself, also towards positive \(x\). Therefore, their electric-field contributions reinforce each other throughout the region between the charges. Consequently, the electric field cannot vanish at the midpoint or anywhere else between the two charges. The student's claim is incorrect. \(\textbf{Identify the possible cancellation region:}\) Outside the two charges, their electric-field contributions oppose each other. For two unequal charges of opposite signs, the finite zero-field point lies on the exterior side of the charge with smaller magnitude. Here, the smaller charge is \(+Q\), located at the origin. Therefore, the possible region is \[ x\lt0 \] \(\textbf{Express the distances:}\) Let the zero-field point be \[ x=-s,\qquad s\gt0 \] Its distance from \(+Q\) is \(s\). Its distance from \(-4Q\) is \(3a+s\). \(\textbf{Apply the electric-field cancellation condition:}\) The field magnitude due to \(+Q\) is \[ E_1=\frac{kQ}{s^2} \] The field magnitude due to \(-4Q\) is \[ E_2=\frac{4kQ}{(3a+s)^2} \] For zero resultant electric field, the opposing contributions must have equal magnitudes: \[ E_1=E_2 \] Therefore, \[ \frac{kQ}{s^2}=\frac{4kQ}{(3a+s)^2} \] Cancelling the common factor \(kQ\), \[ \frac1{s^2}=\frac4{(3a+s)^2} \] Since both distances are positive, taking positive square roots gives \[ \frac1s=\frac2{3a+s} \] Cross-multiplying, \[ 3a+s=2s \] Hence, \[ s=3a \] The zero-field coordinate is \[ x=-3a \] \(\textbf{Verify the physical result:}\) At \(x=-3a\), the positive source produces a field towards negative \(x\), while the negative source produces a field towards positive \(x\). The distances from the two charges are \(3a\) and \(6a\), respectively. Their charge-magnitude ratio is \(1:4\), and their squared-distance ratio is also \(1:4\). Thus, the electric-field magnitudes are equal, while their directions are opposite. The resultant electric field is therefore zero at this point. \(\textbf{Final conclusion:}\) The student's claim is incorrect because the fields due to opposite charges reinforce each other between the charges. The only finite zero-field point lies to the left of \(+Q\), at \(x=-3a\). Hence, option C is correct.

Explanation: \(\textbf{Interpret the field graph:}\) For a point charge in a homogeneous dielectric, \[ E=\frac{k|Q|}{\varepsilon_r r^2} \] Taking \(1/r^2\) as the horizontal variable, \[ E=\left(\frac{k|Q|}{\varepsilon_r}\right)\left(\frac1{r^2}\right) \] The graph is therefore linear, with slope \[ S=\frac{k|Q|}{\varepsilon_r} \] The slope contains the charge magnitude and the medium's relative permittivity. It does not independently reveal the charge sign. \(\textbf{Determine source }A\textbf{:}\) Source \(A\) is in vacuum, so \[ \varepsilon_{rA}=1 \] Its measured slope is \[ S_A=2kQ_0 \] Therefore, \[ k|Q_A|=2kQ_0 \] \[ |Q_A|=2Q_0 \] The field points away from \(A\), identifying a positive source: \[ Q_A=+2Q_0 \] \(\textbf{Determine source }B\textbf{:}\) The relative permittivity for \(B\) is \(3\). Its measured slope is \[ S_B=3kQ_0 \] Using the slope relation, \[ \frac{k|Q_B|}{3}=3kQ_0 \] Cancelling \(k\), \[ |Q_B|=9Q_0 \] The field points towards source \(B\), so its charge is negative: \[ Q_B=-9Q_0 \] \(\textbf{Compare fields at equal distances:}\) At any common observation distance \(r\), \[ E_A=\frac{k(2Q_0)}{r^2} \] For source \(B\), \[ E_B=\frac{k(9Q_0)}{3r^2} \] \[ E_B=\frac{3kQ_0}{r^2} \] Hence, \[ \frac{E_B}{E_A}=\frac{3kQ_0/r^2}{2kQ_0/r^2} \] \[ \frac{E_B}{E_A}=\frac32 \] \(\textbf{Physical interpretation:}\) Although source \(B\) has a larger charge magnitude, its dielectric medium weakens the electric field. The field-magnitude ratio is determined by the graph slopes, while the source signs require the separate directional observations. \(\textbf{Final answer:}\) \[ \boxed{Q_A=+2Q_0,\qquad Q_B=-9Q_0,\qquad \frac{E_B}{E_A}=\frac32} \]

Explanation: \(\textbf{Determine the source field at each position:}\) The source charge \(+Q\) produces an electric field directed radially outward. Both observation points lie on the positive \(x\)-axis, so the field at both positions points towards positive \(x\). The field magnitude at distance \(r\) is \[ E=\frac{kQ}{r^2} \] At \(x=a\), \[ E_A=\frac{kQ}{a^2} \] At \(x=2a\), \[ E_B=\frac{kQ}{(2a)^2} \] \[ E_B=\frac{kQ}{4a^2}=\frac{E_A}{4} \] \(\textbf{Relate electric field to acceleration:}\) The electric force on a particle is \[ \vec F=q_p\vec E \] Combining this with Newton's second law gives \[ \vec a=\frac{q_p}{m_p}\vec E \] The acceleration magnitude depends on both the local electric-field magnitude and the particle's charge-to-mass ratio. \(\textbf{Acceleration of particle }A\textbf{:}\) Its charge is \(+2q\), and its mass is \(4m\). Therefore, \[ a_A=\frac{2q}{4m}E_A \] Substituting the source field, \[ a_A=\frac{2q}{4m}\frac{kQ}{a^2} \] \[ a_A=\frac{kqQ}{2ma^2} \] The positive charge accelerates towards positive \(x\). \(\textbf{Acceleration of particle }B\textbf{:}\) Its charge magnitude is \(q\), and its mass is \(m\). Thus, \[ a_B=\frac{q}{m}E_B \] \[ a_B=\frac qm\frac{kQ}{4a^2} \] \[ a_B=\frac{kqQ}{4ma^2} \] Particle \(B\) is negative, so its acceleration is opposite to the field, towards negative \(x\). \(\textbf{Compare the accelerations:}\) \[ \frac{a_A}{a_B}= \frac{kqQ/(2ma^2)}{kqQ/(4ma^2)} \] \[ \frac{a_A}{a_B}=2 \] \(\textbf{Physical check:}\) Particle \(B\) has a larger charge-to-mass ratio magnitude, but it experiences a field four times weaker. The two effects combine to give half the acceleration magnitude of \(A\). Their charge signs ensure opposite acceleration directions. \(\textbf{Final answer:}\) \[ \boxed{\frac{a_A}{a_B}=2,\qquad \vec a_A\parallel+\hat i,\quad \vec a_B\parallel-\hat i} \]

Explanation: \(\textbf{Identify the symmetric horizontal contributions:}\) The two charges at \(A\) and \(B\) are identical and positive. They lie at equal distances \(a\) on opposite sides of the origin. The field due to \(A\) points away from \(A\), towards positive \(x\). The field due to \(B\) points away from \(B\), towards negative \(x\). Their magnitudes are equal: \[ E_A=E_B=\frac{kQ}{a^2}=E_0 \] Therefore, \[ \vec E_A=+E_0\hat i \] \[ \vec E_B=-E_0\hat i \] Their resultant is \[ \vec E_A+\vec E_B=\vec0 \] \(\textbf{Field due to the third charge:}\) The third source \(-X\) lies at \((0,a)\). Because it is negative, its electric field at the origin points towards the source, along positive \(y\). The separation is \(a\). Hence, \[ \vec E_C=+\frac{kX}{a^2}\hat j \] The complete resultant is therefore \[ \vec E_{\mathrm{net}}=\frac{kX}{a^2}\hat j \] \(\textbf{Apply the required field magnitude:}\) The prescribed resultant is \[ \vec E_{\mathrm{net}}=+2E_0\hat j \] Therefore, \[ \frac{kX}{a^2}=2E_0 \] Using \(E_0=kQ/a^2\), \[ \frac{kX}{a^2}=2\frac{kQ}{a^2} \] Cancelling common factors, \[ X=2Q \] \(\textbf{Determine the force on the negative test charge:}\) The test charge is \(-q\). Its force is \[ \vec F=(-q)\vec E_{\mathrm{net}} \] Substituting the prescribed resultant, \[ \vec F=(-q)(2E_0\hat j) \] \[ \vec F=-2qE_0\hat j \] \(\textbf{Physical check:}\) The electric field points upwards towards the negative source at \(C\). The negative test charge is repelled by that source and therefore experiences a downward force. The two symmetric positive sources make no net contribution at the origin. \(\textbf{Final answer:}\) \[ \boxed{X=2Q,\qquad \vec F=-2qE_0\hat j} \]

Explanation: Diagram P is inconsistent because an electrostatic field has a unique vector direction at every point where its magnitude is nonzero. The tangent to a field line represents the local electric-field direction. If two lines crossed with distinct tangents, the same point would have two different field directions, which is impossible. Diagram Q is consistent with the accepted convention for an electrostatic field-line pattern. Field lines originate on positive charges and terminate on negative charges, or may extend to or from infinity when appropriate. The line need not be straight because the electric field is the vector sum of contributions from all source charges. Curvature is therefore compatible with electrostatic field lines. Diagram R is inconsistent because electrostatic field lines do not form closed loops. A closed directed loop would represent a field direction returning continuously to its starting point, contrary to the character of an electrostatic field. This distinguishes electrostatic field-line patterns from some magnetic-field patterns. Diagram S is consistent because the field of an isolated positive point charge is directed radially outward. Its field lines can extend to infinity in the absence of negative charges on which they terminate. Such radial lines do not intersect except at the source point where the idealised point charge is located and the field is not finite. Consequently, only Q and S are physically acceptable descriptions. The number of lines drawn is a representational choice; their directions and allowed topology must remain physically consistent.

Explanation: \(\textbf{Interpret the field-line density convention:}\) In a consistently drawn field-line representation, greater line density indicates greater electric-field magnitude. The same convention and transverse scale are explicitly used for both regions. Thus, the field magnitudes are proportional to the stated transverse line densities. For region \(A\), \[ n_A=6 \] For region \(B\), \[ n_B=2 \] Therefore, \[ \frac{E_A}{E_B}=\frac{n_A}{n_B} \] \[ \frac{E_A}{E_B}=\frac62=3 \] \(\textbf{Determine the force in region }A\textbf{:}\) The charge is \(+q\). Its force magnitude is \[ F_A=qE_A \] The charge is positive, so its force is along the field direction. Hence, it experiences force towards positive \(x\). \(\textbf{Determine the force in region }B\textbf{:}\) The charge is \(-3q\). Its electric-force magnitude is \[ F_B=3qE_B \] The field in region \(B\) points towards negative \(x\). A negative charge experiences force opposite to the electric field. Therefore, this force also points towards positive \(x\). \(\textbf{Compare the force magnitudes:}\) \[ \frac{F_A}{F_B}=\frac{qE_A}{3qE_B} \] \[ \frac{F_A}{F_B}=\frac13\frac{E_A}{E_B} \] Substituting the field ratio, \[ \frac{F_A}{F_B}=\frac13(3) \] \[ \frac{F_A}{F_B}=1 \] \(\textbf{Interpret the result:}\) The electric field in region \(A\) is three times stronger. However, the magnitude of the test charge in region \(B\) is three times larger. These effects compensate, giving equal force magnitudes. The opposite field arrows do not imply opposite forces because the test charges have opposite signs. \(\textbf{Final answer:}\) \[ \boxed{\frac{F_A}{F_B}=1,\qquad \vec F_A,\vec F_B\parallel+\hat i} \]

Explanation: \(\textbf{Translate the drawing convention:}\) The illustrator represents each positive charge-magnitude unit \(Q\) by four lines. The positive source carries charge \(+3Q\). Therefore, its assigned number of emerging schematic lines is \[ N_+=3\times4=12 \] The negative source has charge magnitude \(Q\). It can therefore be assigned four terminating lines: \[ N_-=1\times4=4 \] \(\textbf{Use the field-line direction rule:}\) Electric-field lines are directed away from positive charges and towards negative charges. Consequently, lines must not be drawn emerging from the negative source and terminating on the positive one. The arrows have physical significance because they indicate the direction of the field acting on a positive test charge. \(\textbf{Account for the unequal source magnitudes:}\) The positive charge has three times the magnitude of the negative charge. Thus, within the chosen convention, more lines originate from the positive charge than can terminate on the negative one. The number of lines ending on the negative source is \[ N_{\mathrm{terminate}}=4 \] The remaining lines extend to infinity: \[ N_{\infty}=N_+-N_{\mathrm{terminate}} \] \[ N_{\infty}=12-4 \] \[ N_{\infty}=8 \] \(\textbf{Check the complete pattern:}\) The total number of lines emerging from \(+3Q\) is twelve. Four terminate on \(-Q\), while the remaining eight extend to infinity. This is compatible with the positive net charge of the isolated pair. The field-line count is only a graphical representation proportional to charge magnitude; electric-field lines are not discrete physical threads. What matters is the consistent convention, the permitted endpoints and the correct arrow directions. \(\textbf{Final answer:}\) \[ \boxed{N_+=12,\qquad N_{\mathrm{terminate}}=4,\qquad N_{\infty}=8} \]

Explanation: \(\textbf{Identify field direction and electric force:}\) The uniform electric field is \[ \vec E=E\hat i \] The proton carries positive charge \(+e\). Therefore, its electric force is \[ \vec F=e\vec E \] \[ \vec F=eE\hat i \] The electric force acts only along positive \(x\). \(\textbf{Determine acceleration components:}\) Using Newton's second law, \[ \vec a=\frac{\vec F}{m} \] \[ \vec a=\frac{eE}{m}\hat i \] Therefore, \[ a_x=\frac{eE}{m},\qquad a_y=0 \] \(\textbf{Use the initial-velocity components:}\) The proton enters with velocity \(v\hat j\). Hence, \[ u_x=0,\qquad u_y=v \] Its initial horizontal velocity is zero, but its vertical velocity is nonzero. \(\textbf{Calculate horizontal displacement:}\) With constant horizontal acceleration, \[ x=u_xt+\frac12a_xt^2 \] Substituting, \[ x=0+\frac12\left(\frac{eE}{m}\right)t^2 \] \[ x=\frac{eEt^2}{2m} \] \(\textbf{Calculate vertical displacement:}\) There is no vertical electric force, so vertical velocity remains constant. \[ y=u_yt+\frac12a_yt^2 \] \[ y=vt \] \(\textbf{Identify the trajectory:}\) Eliminating time using \[ t=\frac yv \] gives \[ x=\frac{eE}{2m}\left(\frac yv\right)^2 \] \[ x=\frac{eE}{2mv^2}y^2 \] This is a parabolic relation between \(x\) and \(y\). \(\textbf{Distinguish trajectory from field line:}\) An electric-field line specifies the direction of the electric field, and hence the electric force on a positive charge. It does not necessarily specify the path of a moving charged particle. The proton begins with vertical velocity even though the electric force is horizontal. Its trajectory therefore curves rather than coinciding with a horizontal field line. \(\textbf{Final answer:}\) \[ \boxed{x=\frac{eEt^2}{2m},\qquad y=vt,\qquad x=\frac{eE}{2mv^2}y^2} \]

Explanation: An electric-field line is a curve whose tangent at each point gives the local direction of the electric field. The direction can change from one point to another because an electrostatic field need not be spatially uniform. Therefore, a smooth line may have a tangent towards positive \(x\) at one point, positive \(y\) at another, and negative \(x\) at a third. This does not violate the field-line rules provided the line does not intersect another line in a region of nonzero field or form a closed electrostatic loop. At point P, the tangent indicates that the electric field points towards positive \(x\). The electric force on a charge is \(\vec F=q\vec E\). For a negative charge, the force direction is opposite to the local electric-field direction. Thus, a negative charge placed at P experiences a force towards negative \(x\). If the charge is initially at rest, its initial velocity is zero, but that does not make the force or acceleration zero. By Newton's second law, its initial acceleration is in the direction of the resultant force. Consequently, the charge initially accelerates towards negative \(x\). The subsequent trajectory cannot be inferred from the field-line tangent alone because it depends on the particle's velocity as well as the spatial variation of the force. The diagram's tangent information is consistent with electrostatics, but the accompanying acceleration claim is incorrect.

Explanation: The Assertion correctly states a fundamental property of electrostatic field-line representations. At a point where the resultant electric field is nonzero, the field can be represented by one definite vector. That vector has a unique direction. A field line is drawn so that its tangent at each point agrees with the direction of the electric field there. Suppose two distinct lines intersected at a point and possessed different tangents at the intersection. The diagram would then assign two different directions to the electric field at one point. This is incompatible with the vector nature of the electric field. Therefore, the two lines cannot cross in that manner. The Reason correctly identifies the physical basis of the nonintersection rule. It is not merely a convention intended to make drawings aesthetically clear. The rule follows from the uniqueness of the resultant field direction at a point of nonzero field. If the field magnitude is zero, no field direction is defined there, and the familiar tangent-direction argument must not be applied as though a nonzero vector were present. The Assertion explicitly refers to a nonzero field, so this qualification does not undermine its validity. Both statements are true, and the Reason correctly explains the Assertion.

Explanation: Uniform electric fields are represented by straight, parallel, equally spaced field lines. Straightness indicates that the local field direction does not change along the represented lines. Parallelism and equal spacing are consistent with a uniform field direction and magnitude throughout the region. In a field-line diagram drawn using a fixed convention, the density of lines represents field strength. Increasing the field magnitude by a factor of two therefore requires twice the line density under the same drawing scale and convention. If the original diagram shows five lines crossing a specified transverse width, the modified representation should show ten across that same width. The source modification also reverses the electric-field direction. This must be represented by reversing the arrows so that they point towards negative \(x\). Merely increasing the drawn lengths of the lines does not represent a stronger electric field. Retaining the original arrow directions would incorrectly indicate that the field still points towards positive \(x\). Curving or converging the field lines would represent a changing direction or magnitude, contrary to the specified uniform field. Thus, both the arrow direction and line density must change, while the lines remain straight, parallel and equally spaced.

Explanation: The original configuration contains two identical positive source charges at symmetric positions about the origin. Their electric-field contributions at the midpoint have equal magnitudes and opposite directions, so the resultant field there is zero. The field-line pattern is correspondingly symmetric. Because both sources are positive, field lines originate from them and extend outward; they do not run from one positive source to the other. Now reverse the signs of both charges while leaving their magnitudes and positions unchanged. At every observation point other than the charge positions, the field due to each source reverses direction because the signed source charge changes sign. The magnitudes of the individual contributions remain unchanged. Their vector sum consequently reverses wherever the original resultant was nonzero. Therefore, the geometrical shapes of the field lines can remain the same, provided every arrow is reversed. The lines now terminate on the negative source charges, having come from infinity where appropriate. At the midpoint, the original vector sum was zero, and reversing each contribution still gives zero. Thus, the zero-field location is preserved. No field line needs to originate at the midpoint, and negative charges are not joined by field lines in the manner proposed by the incorrect descriptions. The transformation changes field direction globally but preserves field magnitudes and the symmetric null point.

Explanation: The two source charges have equal magnitudes and opposite signs. At the midpoint, the field due to the positive charge points away from it, towards the negative charge. The field due to the negative charge points towards that charge, again in the same direction. Thus, the two contributions reinforce rather than cancel. The midpoint field is therefore nonzero. Now consider another finite point on the perpendicular bisector. The distances from that point to the two sources are equal, so the two individual field magnitudes are equal. However, their directions are not exactly opposite. The components perpendicular to the line joining the charges cancel because of geometrical symmetry. Their components parallel to the line joining the charges point from the positive source towards the negative source and reinforce. Consequently, the resultant field remains directed from \(+Q\) towards \(-Q\). Its magnitude generally differs from the midpoint magnitude because the distance and component geometry change. The fact that the total source charge is zero does not imply that the field vanishes. Nor does equality of the distances to the two sources guarantee cancellation. Vector cancellation requires both equal magnitudes and opposite directions. The field-line description correctly reflects this reinforcement along the perpendicular bisector.

Explanation: Statement I is correct because the tangent to the field line at \(P\) identifies the local electric-field direction as positive \(x\). A negative particle experiences electric force opposite to the field, so its force points towards negative \(x\). Statement II is correct for the same physical reason. At \(R\), the electric field points towards positive \(y\), making the force on the negative charge directed towards negative \(y\). Statement III is incorrect because a field line represents the direction of the electric force per unit positive charge, not necessarily the instantaneous velocity of a particle. A particle may possess velocity in a direction different from the local electric force. Its acceleration is determined by the force, while its velocity evolves according to its previous motion as well as the applied forces. Consequently, charged-particle trajectories need not coincide with field lines. Statement IV is correct because tangent direction supplies directional information but not, by itself, the magnitude of the electric field. Field magnitude must be determined from additional quantitative information or from a consistently drawn field-line-density representation. A line may curve between \(P\) and \(R\), so the change in tangent direction does not by itself imply that the field magnitude has changed. These distinctions prevent the common confusion between field direction, force direction, acceleration direction and particle trajectory.