Electrostatics MCQs with Answers – Part 3 (NEET Physics)

Explanation: \(\textbf{Determine the source dipole's field at the probe position:}\) Dipole \(A\) is centred at the origin and has moment \[ \vec p_A=p\hat i \] The observation point \(P(0,r)\) lies on its equatorial line. Because \(r\) is much larger than the source dipole separation, its equatorial field is approximately \[ \vec E_A(P)\simeq-\frac{kp}{r^3}\hat i \] The negative sign indicates that the equatorial field points opposite to the source moment. \(\textbf{Identify the probe dipole moment:}\) The second dipole initially has moment \[ \vec p_B=p_B\hat j \] It points vertically upwards. The local field at its midpoint points towards negative \(x\). Therefore, the moment and the local field are perpendicular. \(\textbf{Apply the dipole torque relation:}\) Since dipole \(B\) is sufficiently small compared with the scale over which the source field changes, the local-field torque approximation is \[ \vec\tau_B\simeq\vec p_B\times\vec E_A(P) \] Substituting, \[ \vec\tau_B\simeq(p_B\hat j)\times\left(-\frac{kp}{r^3}\hat i\right) \] Using \[ \hat j\times\hat i=-\hat k \] we obtain \[ \vec\tau_B\simeq-\frac{kpp_B}{r^3}(-\hat k) \] \[ \vec\tau_B\simeq+\frac{kpp_B}{r^3}\hat k \] \(\textbf{Interpret the rotational tendency:}\) Positive \(\hat k\) corresponds to an anticlockwise torque when viewed from positive \(z\). The probe moment initially points along positive \(y\). An anticlockwise rotation from positive \(y\) moves the moment towards negative \(x\), which is the direction of the local field. Thus, the torque tends to align dipole \(B\) with the field created by dipole \(A\). \(\textbf{Physical qualification:}\) The field of dipole \(A\) is not globally uniform. However, the probe dipole is much smaller than its distance from \(A\), making the local uniform-field approximation appropriate for calculating the leading torque about the probe midpoint. \(\textbf{Final answer:}\) \[ \boxed{\vec\tau_B\simeq+\frac{kpp_B}{r^3}\hat k\quad\text{(anticlockwise towards }-\hat i\text{)}} \]

Explanation: \(\textbf{Initial dipole moment and midpoint field:}\) The initial charges are \[ q_A=-5e,\qquad q_B=+5e \] Their separation is \(2a\). The dipole moment points from negative to positive charge: \[ \vec p_i=(5e)(2a)\hat i \] \[ \vec p_i=+10ea\hat i \] At the midpoint, the fields due to both charges point towards negative \(x\). Each field has magnitude \[ E_1=\frac{k(5e)}{a^2} \] Therefore, \[ \vec E_i=-\frac{10ke}{a^2}\hat i \] \(\textbf{Account for the electron transfer:}\) Sphere A loses \(n\) electrons and becomes more positively charged. Sphere B gains the same number and becomes less positively charged. Their final charges are \[ q_A'=(-5+n)e \] \[ q_B'=(5-n)e \] Their sum remains zero: \[ q_A'+q_B'=0 \] \(\textbf{Apply the field-reversal condition:}\) The midpoint field reverses only when the charge signs have interchanged. Thus, \[ n\gt5 \] The left sphere is then positive, and the right sphere negative. Both midpoint-field contributions point towards positive \(x\). Their resultant magnitude is \[ E_f=\frac{2k(n-5)e}{a^2} \] The measured magnitude ratio is \[ \frac{E_f}{E_i}=\frac35 \] Substituting, \[ \frac{2k(n-5)e/a^2}{10ke/a^2}=\frac35 \] \[ \frac{n-5}{5}=\frac35 \] \[ n-5=3 \] \[ n=8 \] \(\textbf{Determine the final dipole moment:}\) The final charges become \[ q_A'=+3e,\qquad q_B'=-3e \] The dipole moment now points from B towards A, along negative \(x\). Its magnitude is \[ p_f=(3e)(2a)=6ea \] Therefore, \[ \vec p_f=-6ea\hat i \] \(\textbf{Physical verification:}\) The initial midpoint field magnitude was proportional to \(5e\), while the final magnitude is proportional to \(3e\). Its direction reverses because the positions of the positive and negative charges have exchanged. The electron count is an integer, consistent with charge quantisation. \(\textbf{Final answer:}\) \[ \boxed{n=8,\qquad \vec p_f=-6ea\hat i} \]

Explanation: \(\textbf{Use the mutual Coulomb force:}\) The particles have equal charge magnitudes and opposite signs. Their attractive force magnitude is \[ F=\frac{kq^2}{d^2} \] The measurement gives \[ \frac{kq^2}{d^2}=\frac{kq_0^2}{a^2} \] Cancelling \(k\), \[ \frac{q^2}{d^2}=\frac{q_0^2}{a^2} \] Both charge magnitudes and distances are positive, so \[ \frac qd=\frac{q_0}{a} \] \(\textbf{Use the perpendicular-field torque:}\) The electric dipole moment magnitude is \[ p=qd \] The torque in a uniform field is \[ \tau=pE\sin\theta \] The dipole is perpendicular to the field: \[ \theta=90^\circ \] Therefore, \[ \tau=qdE \] The measured torque is \(4q_0aE\), giving \[ qdE=4q_0aE \] Cancelling \(E\), \[ qd=4q_0a \] \(\textbf{Solve the two independent relations:}\) From the Coulomb-force measurement, \[ q=\frac{q_0}{a}d \] Substitute this into the torque relation: \[ \left(\frac{q_0d}{a}\right)d=4q_0a \] \[ \frac{q_0d^2}{a}=4q_0a \] \[ d^2=4a^2 \] Since \(d\gt0\), \[ d=2a \] Then, \[ q=\frac{q_0}{a}(2a) \] \[ q=2q_0 \] \(\textbf{Check both measurements:}\) The predicted mutual force is \[ F=\frac{k(2q_0)^2}{(2a)^2}=\frac{kq_0^2}{a^2} \] The predicted maximum torque is \[ \tau=(2q_0)(2a)E=4q_0aE \] Both agree with the observations. \(\textbf{Final answer:}\) \[ \boxed{q=2q_0,\qquad d=2a} \]

Explanation: \(\textbf{Establish the original combined moment:}\) Electric dipole moments add vectorially. Initially, \[ \vec p_i=\vec p_A+\vec p_B \] \[ \vec p_i=p_0(3\hat i+4\hat j) \] The applied field is \[ \vec E=\frac E5(3\hat i+4\hat j) \] The original combined moment is parallel to the field. Therefore, \[ \vec\tau_i=\vec p_i\times\vec E=\vec0 \] This verifies the stated initial zero torque. \(\textbf{Reverse the charges of dipole B:}\) Reversing both charge signs while keeping their positions fixed reverses the dipole moment vector. Thus, \[ \vec p_B'=-4p_0\hat j \] Dipole A remains unchanged: \[ \vec p_A'=3p_0\hat i \] The modified combined moment is \[ \vec p_f=p_0(3\hat i-4\hat j) \] \(\textbf{Determine the resultant force:}\) The applied field remains uniform. For every constituent dipole, the forces on its equal and opposite charges cancel. Therefore, the entire frame has zero resultant electric force: \[ \vec F_{\mathrm{net}}=\vec0 \] \(\textbf{Calculate the modified torque:}\) The torque on the frame is \[ \vec\tau_f=\vec p_f\times\vec E \] For vectors in the \(xy\)-plane, \[ \tau_z=p_xE_y-p_yE_x \] Here, \[ p_x=3p_0,\qquad p_y=-4p_0 \] and \[ E_x=\frac{3E}{5},\qquad E_y=\frac{4E}{5} \] Therefore, \[ \tau_z=(3p_0)\left(\frac{4E}{5}\right)-(-4p_0)\left(\frac{3E}{5}\right) \] \[ \tau_z=\frac{12}{5}p_0E+\frac{12}{5}p_0E \] \[ \tau_z=\frac{24}{5}p_0E \] Hence, \[ \vec\tau_f=+\frac{24}{5}p_0E\hat k \] \(\textbf{Physical interpretation:}\) The initial zero torque results from alignment of the combined moment with the field. Reversing only one constituent moment destroys that alignment. The frame can therefore experience a nonzero torque while its net electric force remains zero. \(\textbf{Final answer:}\) \[ \boxed{\vec F_{\mathrm{net}}=\vec0,\qquad \vec\tau_f=+\frac{24}{5}p_0E\hat k} \]

Explanation: \(\textbf{Initial dipole properties:}\) The negative charge lies at \((-a,0)\), and the positive charge lies at \((a,0)\). The electric dipole moment is \[ \vec p_i=2qa\hat i \] The applied uniform field is \[ \vec E=E\hat j \] Therefore, the initial torque is \[ \vec\tau_i=\vec p_i\times\vec E \] \[ \vec\tau_i=(2qa\hat i)\times(E\hat j) \] \[ \vec\tau_i=+2qaE\hat k \] \(\textbf{Initial field at the midpoint:}\) The positive sphere produces a field towards negative \(x\) at the midpoint. The negative sphere also produces a field towards negative \(x\). Each contribution has magnitude \[ E_1=\frac{kq}{a^2} \] Thus, the field due to the spheres is \[ \vec E_{\mathrm{dipole},i}=-\frac{2kq}{a^2}\hat i \] \(\textbf{Redistribution during contact:}\) The spheres are identical conductors. Their total charge is \[ Q_{\mathrm{total}}=-q+q=0 \] On contact, charge redistributes until the identical spheres attain the same charge. The final charge on each sphere is \[ q_f=\frac{Q_{\mathrm{total}}}{2} \] \[ q_f=0 \] \(\textbf{Properties after returning the spheres:}\) Both spheres are now neutral in the ideal point-charge model. They no longer constitute the original pair of equal and opposite point charges. Their resulting dipole moment is \[ \vec p_f=\vec0 \] They produce no point-charge electric field at the midpoint: \[ \vec E_{\mathrm{dipole},f}=\vec0 \] The uniform external field may still exist, but the neutralised point-charge pair experiences no dipole torque: \[ \vec\tau_f=\vec0 \] \(\textbf{Physical distinction:}\) Total charge was zero both before and after contact. However, the initial system had separated opposite charges, whereas the final idealised system has no net charge on either sphere. Charge conservation does not require the dipole moment to be conserved. The correct comparison is therefore the one stated in option D.

Explanation: Statement I is correct because the two constituent charges have equal magnitudes and opposite signs. Under one consistent line-count convention, the number of lines assigned to the positive charge can equal the number terminating on the negative charge. The direction of these lines is from positive to negative. Statement II is correct because the electric dipole moment points from the negative charge towards the positive charge. In the stated arrangement, the dipole moment points towards the right. At a point on the perpendicular bisector, the components of the two charge fields perpendicular to the dipole axis cancel, while the components directed from positive to negative reinforce. Therefore, the resultant field points towards the left, opposite to the dipole moment. Statement III is incorrect because the electric fields of the two charges do not cancel at the midpoint. The positive charge produces a field directed away from itself, towards the negative charge. The negative charge produces a field directed towards itself, in that same direction. The contributions reinforce, even though the algebraic total charge is zero. Statement IV is incorrect because electrostatic field lines do not form closed loops. Their directions are determined by the electrostatic field, and the lines do not return to the positive charge from which they originate. Thus, Statements I and II are correct. The question distinguishes charge neutrality from field cancellation and relates the dipole moment direction to the field-line pattern.

Explanation: \(\textbf{Recognise the field-distance behaviour:}\) A distant axial electric field due to a small dipole has magnitude \[ E_{\mathrm{axial}}\simeq\frac{2kp}{r^3} \] The dipole moment magnitude is \[ p=qd \] Therefore, \[ E_{\mathrm{axial}}\simeq\frac{2kqd}{r^3} \] The inverse-cube distance dependence predicts \[ \frac{E(2r)}{E(r)}=\frac18 \] The measured force ratio is consistent with this behaviour because the same probe charge is used in both measurements. \(\textbf{Relate the first force measurement to the dipole field:}\) The positive probe has magnitude \(q_0\). Its force magnitude at distance \(r\) is \[ F_0=q_0E(r) \] Substituting the distant axial field, \[ F_0=q_0\frac{2kqd}{r^3} \] \[ F_0=\frac{2kqq_0d}{r^3} \] \(\textbf{Infer the constituent charge magnitude:}\) Rearranging, \[ F_0r^3=2kqq_0d \] Therefore, \[ q=\frac{F_0r^3}{2kq_0d} \] \(\textbf{Predict the force at }3r\textbf{:}\) Within the same distant-dipole approximation, \[ F(3r)=q_0\frac{2kp}{(3r)^3} \] Substituting \(p=qd\), \[ F(3r)=\frac{2kqq_0d}{27r^3} \] Using \[ F_0=\frac{2kqq_0d}{r^3} \] gives \[ F(3r)=\frac{F_0}{27} \] \(\textbf{Physical check:}\) An isolated point charge would produce a field proportional to \(1/r^2\). Here, the observed one-eighth reduction on doubling distance is consistent with the leading \(1/r^3\) dipole field. The inferred charge is a positive magnitude; the dipole contains charges \(+q\) and \(-q\). \(\textbf{Final answer:}\) \[ \boxed{q=\frac{F_0r^3}{2kq_0d},\qquad F(3r)\simeq\frac{F_0}{27}} \]

Explanation: \(\textbf{Determine the distant axial field at P:}\) The dipole moment points along positive \(x\). Point P lies on the positive axial side of the dipole. Its field is directed along the moment: \[ \vec E_P\simeq+\frac{2kp}{r^3}\hat i \] \(\textbf{Calculate the force on the negative probe:}\) The probe charge at P is \[ q_P=-q \] Therefore, \[ \vec F_P=q_P\vec E_P \] \[ \vec F_P=(-q)\left(\frac{2kp}{r^3}\hat i\right) \] \[ \vec F_P=-\frac{2kpq}{r^3}\hat i \] This force points towards negative \(x\). \(\textbf{Determine the equatorial field at R:}\) Point R lies on the perpendicular bisector of the dipole. The distant equatorial field points opposite to the moment and has magnitude \(kp/r^3\). Thus, \[ \vec E_R\simeq-\frac{kp}{r^3}\hat i \] \(\textbf{Impose equality of force vectors:}\) For the second probe charge \(q_R\), \[ \vec F_R=q_R\vec E_R \] Therefore, \[ \vec F_R=-\frac{kpq_R}{r^3}\hat i \] The requirement is \[ \vec F_R=\vec F_P \] Hence, \[ -\frac{kpq_R}{r^3}\hat i=-\frac{2kpq}{r^3}\hat i \] Cancelling common factors, \[ q_R=2q \] \(\textbf{Direction and magnitude check:}\) The field at R is half as strong as the field at P, but it points in the opposite direction. A positive probe at R therefore experiences force towards negative \(x\), just like the negative probe at P. Doubling its charge magnitude compensates for the weaker equatorial field. \(\textbf{Final answer:}\) \[ \boxed{q_R=+2q,\qquad \vec F_R=\vec F_P=-\frac{2kpq}{r^3}\hat i} \]

Explanation: Statement I is correct because the initial pair contains equal and opposite charges separated by distance \(2a\). Its algebraic total charge is zero, but its dipole moment is nonzero. At sufficiently large distances on the axial line, the leading field magnitude is approximately \(2kp/r^3\), where \(p=2qa\). Thus, the initial distant field has inverse-cube dependence. Statement II is correct because reversing the positive charge converts the pair into two negative charges. Their combined charge is now \(-2q\), which is nonzero. At distances much larger than their separation, the leading electric field is then the monopole field of the combined charge, with magnitude approximately \(2kq/r^2\). The leading distance dependence therefore changes from inverse cube to inverse square. Statement III is incorrect because the original exterior axial field points along the initial dipole moment, towards positive \(x\). After the change, both sources are negative. At a distant point on the positive \(x\)-axis, both produce fields directed towards the negative charges, along negative \(x\). The distant-field direction therefore reverses. Statement IV is incorrect because the modified pair is no longer an electric dipole consisting of equal and opposite charges. The original dipole approximation is not the correct leading description of its distant field. The comparison shows that changing one constituent charge sign changes the multipole character of the source arrangement, not merely the magnitude of its existing dipole field. Statements I and II are correct.

Explanation: \(\textbf{Identify the relevant angle:}\) Electric flux through a plane surface in a uniform electric field is \[ \Phi_E=\vec E\cdot\vec A \] The area vector is perpendicular to the plane. Therefore, \[ \Phi_E=EA\cos\theta \] where \(\theta\) is the angle between the electric field and the chosen area vector. \(\textbf{Convert the given plane angle:}\) The electric field makes an angle of \(30^\circ\) with the plane. The corresponding acute angle with the normal is \[ \theta=90^\circ-30^\circ \] \[ \theta=60^\circ \] The chosen normal has a positive component along the field, so \[ \vec E\cdot\vec A\gt0 \] \(\textbf{Evaluate the flux:}\) \[ \Phi_E=EA\cos60^\circ \] Using \[ \cos60^\circ=\frac12 \] we obtain \[ \Phi_E=\frac12EA \] \(\textbf{Physical check:}\) If the field were parallel to the plane, the flux would be zero. If it were perpendicular to the plane and aligned with the area vector, the flux would be \(EA\). The given orientation lies between those limits, giving a positive intermediate value. \(\textbf{Final answer:}\) \[ \boxed{\Phi_E=+\frac12EA} \]

Explanation: \(\textbf{Construct the oriented area vector:}\) The surface area is \[ A=0.20\,\text{m}^2 \] The chosen unit normal is \[ \hat n=\frac35\hat i+\frac45\hat j \] Therefore, \[ \vec A=A\hat n \] \[ \vec A=0.20\left(\frac35\hat i+\frac45\hat j\right) \] \[ \vec A=(0.12\hat i+0.16\hat j)\,\text{m}^2 \] \(\textbf{Calculate the initial signed flux:}\) Electric flux is the scalar product \[ \Phi_i=\vec E\cdot\vec A \] Substituting, \[ \Phi_i=(300\hat i-400\hat j)\cdot(0.12\hat i+0.16\hat j) \] \[ \Phi_i=(300)(0.12)+(-400)(0.16) \] \[ \Phi_i=36-64 \] \[ \Phi_i=-28\,\text{N m}^2\text{C}^{-1} \] The negative value means the field has a net component opposite to the chosen area vector. \(\textbf{Reverse the surface orientation:}\) Turning the oriented surface over reverses its area vector: \[ \vec A'=-\vec A \] The field remains unchanged. Therefore, \[ \Phi_f=\vec E\cdot\vec A' \] \[ \Phi_f=\vec E\cdot(-\vec A) \] \[ \Phi_f=-\Phi_i \] \[ \Phi_f=+28\,\text{N m}^2\text{C}^{-1} \] \(\textbf{Physical interpretation:}\) Reversing the chosen normal changes the sign of flux but not its magnitude. It does not change the electric field or the geometrical area of the surface. The result follows from the orientation dependence of the scalar product. \(\textbf{Final answer:}\) \[ \boxed{\Phi_i=-28\,\text{N m}^2\text{C}^{-1},\qquad \Phi_f=+28\,\text{N m}^2\text{C}^{-1}} \]

Explanation: \(\textbf{Flux in the first experiment:}\) The electric flux through a plane surface in a uniform field is \[ \Phi=EA\cos\theta \] Initially, \[ \theta_I=60^\circ \] Therefore, \[ \Phi_I=EA\cos60^\circ \] \[ \Phi_I=\frac12EA \] The first flux is positive because the chosen area vector makes an acute angle with the field. \(\textbf{Flux after changing area and orientation:}\) In experiment II, the area becomes \[ A_{II}=2A \] The new angle with the field is \[ \theta_{II}=120^\circ \] Thus, \[ \Phi_{II}=E(2A)\cos120^\circ \] Since \[ \cos120^\circ=-\frac12 \] we obtain \[ \Phi_{II}=2EA\left(-\frac12\right) \] \[ \Phi_{II}=-EA \] \(\textbf{Compare the signed fluxes:}\) \[ \frac{\Phi_{II}}{\Phi_I}=\frac{-EA}{EA/2} \] \[ \frac{\Phi_{II}}{\Phi_I}=-2 \] \(\textbf{Physical check:}\) The area doubles, which doubles the magnitude for an otherwise unchanged angle factor. The change from \(60^\circ\) to \(120^\circ\) reverses the sign of the cosine without changing its absolute value. Thus, the final flux is twice as large in magnitude but opposite in sign. \(\textbf{Final answer:}\) \[ \boxed{\frac{\Phi_{II}}{\Phi_I}=-2} \]

Explanation: \(\textbf{Express the first signed flux:}\) The initial angle is acute, so \[ 0^\circ\lt\theta\lt90^\circ \] The initial flux is \[ \Phi_1=EA\cos\theta \] Since \(\cos\theta\gt0\), \[ \Phi_1\gt0 \] \(\textbf{Express the rotated flux:}\) The rotation increases the angle with the field by \(90^\circ\). Therefore, \[ \Phi_2=EA\cos(\theta+90^\circ) \] Using \[ \cos(\theta+90^\circ)=-\sin\theta \] we obtain \[ \Phi_2=-EA\sin\theta \] Because \(\theta\) is acute, \[ \Phi_2\lt0 \] Thus, the two measured fluxes have opposite signs. \(\textbf{Apply the measured magnitude relation:}\) The experiment gives \[ \Phi_1=3|\Phi_2| \] Substituting, \[ EA\cos\theta=3EA\sin\theta \] Cancelling the nonzero factor \(EA\), \[ \cos\theta=3\sin\theta \] Dividing by \(\cos\theta\), \[ 1=3\tan\theta \] \[ \tan\theta=\frac13 \] Therefore, \[ \theta=\tan^{-1}\left(\frac13\right) \] \(\textbf{Determine the signed flux ratio:}\) \[ \frac{\Phi_2}{\Phi_1}=\frac{-EA\sin\theta}{EA\cos\theta} \] \[ \frac{\Phi_2}{\Phi_1}=-\tan\theta \] \[ \frac{\Phi_2}{\Phi_1}=-\frac13 \] \(\textbf{Physical interpretation:}\) The same surface and field are used throughout, so the flux change comes entirely from the orientation. The magnitude condition determines the initial acute angle, while the rotation determines the sign reversal. \(\textbf{Final answer:}\) \[ \boxed{\theta=\tan^{-1}\left(\frac13\right),\qquad \frac{\Phi_2}{\Phi_1}=-\frac13} \]

Explanation: \(\textbf{Identify the area vectors:}\) Every face of the cube has area \[ A=a^2 \] The six outward area vectors are \[ +a^2\hat i,\quad -a^2\hat i \] \[ +a^2\hat j,\quad -a^2\hat j \] \[ +a^2\hat k,\quad -a^2\hat k \] The electric field is uniform throughout the cube. \(\textbf{Calculate the flux through the }+\hat j\textbf{ face:}\) The relevant area vector is \[ \vec A_{+y}=a^2\hat j \] The electric flux is \[ \Phi_{+y}=\vec E\cdot\vec A_{+y} \] Substituting, \[ \Phi_{+y}=E_0(2\hat i-3\hat j+\hat k)\cdot(a^2\hat j) \] Only the \(y\)-component contributes: \[ \Phi_{+y}=-3E_0a^2 \] The flux is negative because the field crosses that face opposite to its outward normal. \(\textbf{Compare opposite faces:}\) For the pair perpendicular to the \(x\)-axis, \[ \Phi_{+x}=+2E_0a^2 \] \[ \Phi_{-x}=-2E_0a^2 \] For the \(y\)-faces, \[ \Phi_{+y}=-3E_0a^2 \] \[ \Phi_{-y}=+3E_0a^2 \] For the \(z\)-faces, \[ \Phi_{+z}=+E_0a^2 \] \[ \Phi_{-z}=-E_0a^2 \] \(\textbf{Add the signed fluxes:}\) The net closed-surface flux is \[ \Phi_{\mathrm{net}}=\Phi_{+x}+\Phi_{-x}+\Phi_{+y}+\Phi_{-y}+\Phi_{+z}+\Phi_{-z} \] Substituting, \[ \Phi_{\mathrm{net}}=2E_0a^2-2E_0a^2-3E_0a^2+3E_0a^2+E_0a^2-E_0a^2 \] \[ \Phi_{\mathrm{net}}=0 \] \(\textbf{Physical interpretation:}\) A uniform electric field can enter a closed surface through some faces and leave through others. The signed contributions cancel pairwise for opposite parallel faces. Zero net flux does not mean the field vanishes on the surface. \(\textbf{Final answer:}\) \[ \boxed{\Phi_{\mathrm{net}}=0,\qquad \Phi_{+y}=-3E_0a^2} \]

Explanation: \(\textbf{Extract the field components from the cubical measurements:}\) Each cube face has area \[ A=a^2 \] For a uniform field, the outward flux through the \(+\hat i\) face is \[ \Phi_{+x}=E_xa^2 \] The measurement gives \[ E_xa^2=2\Phi_0 \] Therefore, \[ E_x=\frac{2\Phi_0}{a^2} \] Similarly, the \(+\hat j\) face measurement gives \[ E_ya^2=-3\Phi_0 \] \[ E_y=-\frac{3\Phi_0}{a^2} \] The \(+\hat k\) face gives \[ E_za^2=4\Phi_0 \] \[ E_z=\frac{4\Phi_0}{a^2} \] \(\textbf{Construct the inferred electric-field vector:}\) \[ \vec E=E_x\hat i+E_y\hat j+E_z\hat k \] Substituting, \[ \vec E=\frac{\Phi_0}{a^2}(2\hat i-3\hat j+4\hat k) \] \(\textbf{Identify the new area vector:}\) The open surface has area \(a^2\). Its chosen unit normal is \[ \hat n=\frac13(2\hat i+2\hat j+\hat k) \] The vector is a unit vector because \[ |\hat n|=\frac13\sqrt{2^2+2^2+1^2}=1 \] Thus, \[ \vec A=a^2\hat n \] \[ \vec A=\frac{a^2}{3}(2\hat i+2\hat j+\hat k) \] \(\textbf{Calculate the open-surface flux:}\) \[ \Phi=\vec E\cdot\vec A \] Substituting, \[ \Phi=\frac{\Phi_0}{a^2}(2\hat i-3\hat j+4\hat k)\cdot\frac{a^2}{3}(2\hat i+2\hat j+\hat k) \] Cancelling \(a^2\), \[ \Phi=\frac{\Phi_0}{3}[(2)(2)+(-3)(2)+(4)(1)] \] \[ \Phi=\frac{\Phi_0}{3}(4-6+4) \] \[ \Phi=+\frac23\Phi_0 \] \(\textbf{Physical check:}\) The negative \(y\)-component contributes negative flux, while the \(x\)- and \(z\)-components contribute positive flux. The final value is positive because the positive contributions dominate. The same uniform field is used in both experiments; only the surface orientation changes. \(\textbf{Final answer:}\) \[ \boxed{\Phi=+\frac23\Phi_0} \]

Explanation: \(\textbf{Use flux additivity across the sections:}\) The complete surface is divided into three non-overlapping sections. Its total electric flux equals the algebraic sum of the fluxes through those sections: \[ \Phi_{\mathrm{total}}=\Phi_P+\Phi_Q+\Phi_R \] The same chosen normal \(+\hat k\) applies to all three sections. \(\textbf{Flux through section P:}\) Its area vector is \[ \vec A_P=A\hat k \] The local electric field is \[ \vec E_P=E_0(\hat i+2\hat k) \] Therefore, \[ \Phi_P=\vec E_P\cdot\vec A_P \] \[ \Phi_P=E_0(\hat i+2\hat k)\cdot(A\hat k) \] \[ \Phi_P=+2E_0A \] \(\textbf{Flux through section Q:}\) Its area is \(2A\), so \[ \vec A_Q=2A\hat k \] The local field is \[ \vec E_Q=E_0(3\hat i-\hat k) \] Hence, \[ \Phi_Q=\vec E_Q\cdot\vec A_Q \] \[ \Phi_Q=E_0(3\hat i-\hat k)\cdot(2A\hat k) \] \[ \Phi_Q=-2E_0A \] The flux is negative because the normal component of the field is opposite to the chosen surface normal. \(\textbf{Flux through section R:}\) The area vector is \[ \vec A_R=A\hat k \] The local field is \[ \vec E_R=E_0(-2\hat j+3\hat k) \] Therefore, \[ \Phi_R=\vec E_R\cdot\vec A_R \] \[ \Phi_R=E_0(-2\hat j+3\hat k)\cdot(A\hat k) \] \[ \Phi_R=+3E_0A \] \(\textbf{Combine the signed contributions:}\) \[ \Phi_{\mathrm{total}}=2E_0A-2E_0A+3E_0A \] \[ \Phi_{\mathrm{total}}=+3E_0A \] \(\textbf{Physical interpretation:}\) Only the field components perpendicular to the surface contribute to flux. The tangential components, however large, make no contribution to the scalar product. The signed section fluxes must be added rather than their magnitudes. \(\textbf{Final answer:}\) \[ \boxed{\Phi_{\mathrm{total}}=+3E_0A} \]

Explanation: \(\textbf{Determine the initial flux:}\) Electric flux through an oriented plane surface in a uniform field is \[ \Phi=EA\cos\theta \] Initially, \[ \theta_0=60^\circ \] Therefore, \[ \Phi_0=EA\cos60^\circ \] \[ \Phi_0=\frac12EA \] This provides the reference signed flux. \(\textbf{Apply the field-strength and orientation changes:}\) The field magnitude becomes \[ E_1=2E \] The new chosen area vector makes an angle of \(120^\circ\) with the unchanged field direction. Therefore, \[ \Phi_1=E_1A\cos120^\circ \] Substituting, \[ \Phi_1=(2E)A\left(-\frac12\right) \] \[ \Phi_1=-EA \] Using \(EA=2\Phi_0\), \[ \Phi_1=-2\Phi_0 \] \(\textbf{Reverse the electric-field direction:}\) The field magnitude remains \(2E\), but its direction reverses. For a fixed area vector, reversing the field changes the dot product's sign: \[ \Phi_2=(-\vec E_1)\cdot\vec A_1 \] \[ \Phi_2=-\Phi_1 \] Therefore, \[ \Phi_2=+EA \] \[ \Phi_2=+2\Phi_0 \] \(\textbf{Check the final angle:}\) The final field direction is opposite to the field direction used to define the \(120^\circ\) angle. Hence, the final angle between the field and area vector is \[ \theta_2=180^\circ-120^\circ \] \[ \theta_2=60^\circ \] Directly, \[ \Phi_2=(2E)A\cos60^\circ=EA=2\Phi_0 \] Both methods agree. \(\textbf{Physical interpretation:}\) The first change increases the flux magnitude and reverses its sign. The second change reverses only the sign, because field magnitude and surface orientation remain unchanged. \(\textbf{Final answer:}\) \[ \boxed{\Phi_1=-2\Phi_0,\qquad \Phi_2=+2\Phi_0} \]

Explanation: \(\textbf{Identify the box-face orientations:}\) The box has six faces arranged in three parallel opposite pairs. For each pair, the two outward area vectors point in opposite directions. Since the electric field is uniform, the field vector is identical at corresponding points on opposite faces. \(\textbf{Relate the fluxes through opposite faces:}\) For a uniform field and a plane face, \[ \Phi=\vec E\cdot\vec A \] Opposite faces of a rectangular box have equal areas but opposite outward area vectors: \[ \vec A_-=-\vec A_+ \] Therefore, \[ \Phi_-=\vec E\cdot(-\vec A_+) \] \[ \Phi_-=-\Phi_+ \] This applies independently to each face pair. \(\textbf{Determine the }-\hat i\textbf{ face flux:}\) The measured \(+\hat i\) face flux is \[ \Phi_{+x}=+6\Phi_0 \] Therefore, \[ \Phi_{-x}=-6\Phi_0 \] \(\textbf{Determine the }-\hat j\textbf{ face flux:}\) The measured \(+\hat j\) face flux is \[ \Phi_{+y}=-4\Phi_0 \] Consequently, \[ \Phi_{-y}=+4\Phi_0 \] \(\textbf{Determine the }-\hat k\textbf{ face flux:}\) The measured \(+\hat k\) face flux is \[ \Phi_{+z}=+3\Phi_0 \] Therefore, \[ \Phi_{-z}=-3\Phi_0 \] \(\textbf{Calculate the net closed-surface flux:}\) Add the six signed contributions: \[ \Phi_{\mathrm{net}}=\Phi_{+x}+\Phi_{-x}+\Phi_{+y}+\Phi_{-y}+\Phi_{+z}+\Phi_{-z} \] Substituting, \[ \Phi_{\mathrm{net}}=6\Phi_0-6\Phi_0-4\Phi_0+4\Phi_0+3\Phi_0-3\Phi_0 \] \[ \Phi_{\mathrm{net}}=0 \] \(\textbf{Physical interpretation:}\) The different face areas affect the individual field components inferred from the measurements. However, they do not prevent cancellation between opposite faces because each opposite pair has equal area. Uniformity of the field ensures that flux entering the closed box equals flux leaving it. The net flux is therefore zero even though several individual face fluxes are nonzero. \(\textbf{Final answer:}\) \[ \boxed{\Phi_{-x}=-6\Phi_0,\qquad \Phi_{-y}=+4\Phi_0,\qquad \Phi_{-z}=-3\Phi_0,\qquad \Phi_{\mathrm{net}}=0} \]

Explanation: \(\textbf{Identify the field components relative to the cylinder:}\) The prescribed electric field is \[ \vec E=3E_0\hat i+4E_0\hat k \] Its horizontal component is perpendicular to the cylinder's axis. Its vertical component is parallel to the cylinder's axis. The horizontal component can cross the curved surface locally, whereas the vertical component cannot cross it because the curved-surface normals are horizontal. \(\textbf{Calculate the top-face flux:}\) The top face has area vector \[ \vec A_{\mathrm{top}}=\pi R^2\hat k \] Its flux is \[ \Phi_{\mathrm{top}}=\vec E\cdot\vec A_{\mathrm{top}} \] \[ \Phi_{\mathrm{top}}=E_0(3\hat i+4\hat k)\cdot(\pi R^2\hat k) \] \[ \Phi_{\mathrm{top}}=+4\pi R^2E_0 \] \(\textbf{Calculate the bottom-face flux:}\) The bottom face has outward area vector \[ \vec A_{\mathrm{bottom}}=-\pi R^2\hat k \] Therefore, \[ \Phi_{\mathrm{bottom}}=E_0(3\hat i+4\hat k)\cdot(-\pi R^2\hat k) \] \[ \Phi_{\mathrm{bottom}}=-4\pi R^2E_0 \] \(\textbf{Determine the curved-surface contribution:}\) The horizontal electric field enters through one side of the curved surface and leaves through the opposite side. Because the cylinder is symmetric and the field is uniform, these signed contributions cancel. The vertical field is tangent to the curved surface everywhere. Consequently, \[ \Phi_{\mathrm{curved}}=0 \] This does not mean that the horizontal field is tangent to every element of the curved surface. It means that its positive and negative contributions give zero net curved-surface flux. \(\textbf{Remove the top face:}\) The remaining open surface consists of the curved surface and the bottom face. Hence, \[ \Phi_{\mathrm{remaining}}=\Phi_{\mathrm{curved}}+\Phi_{\mathrm{bottom}} \] \[ \Phi_{\mathrm{remaining}}=0-4\pi R^2E_0 \] \[ \Phi_{\mathrm{remaining}}=-4\pi R^2E_0 \] \(\textbf{Physical check:}\) For the original closed cylinder, the three contributions sum to zero, as expected for a uniform field. Removing the top face leaves an uncancelled negative contribution from the bottom face. \(\textbf{Final answer:}\) \[ \boxed{\Phi_{\mathrm{curved}}=0,\qquad \Phi_{\mathrm{remaining}}=-4\pi R^2E_0} \]

Explanation: \(\textbf{Determine the normal-field relation:}\) The normal component varies linearly with \(x\). Its slope is \[ \frac{\Delta E_z}{\Delta x}=\frac{3E_0-E_0}{L} \] \[ \frac{\Delta E_z}{\Delta x}=\frac{2E_0}{L} \] Therefore, \[ E_z(x)=E_0+\frac{2E_0x}{L} \] \(\textbf{Calculate the original flux:}\) The surface has length \(L\) along \(y\). Its flux equals \(L\) multiplied by the area under the \(E_z\)-versus-\(x\) graph. The full graph forms a trapezium. Its average field component is \[ \overline E_z=\frac{E_0+3E_0}{2}=2E_0 \] Hence, \[ \Phi_i=\overline E_zL^2 \] \[ \Phi_i=2E_0L^2 \] \(\textbf{Flux through the left half:}\) At the dividing line, \[ E_z(L/2)=E_0+\frac{2E_0(L/2)}{L} \] \[ E_z(L/2)=2E_0 \] The average normal field over the left half is \[ \overline E_L=\frac{E_0+2E_0}{2} \] \[ \overline E_L=\frac32E_0 \] Its area is \(L^2/2\). Therefore, its original positive flux is \[ \Phi_L=\frac32E_0\left(\frac{L^2}{2}\right) \] \[ \Phi_L=\frac34E_0L^2 \] \(\textbf{Flux through the right half:}\) The field component increases from \(2E_0\) to \(3E_0\). Its average is \[ \overline E_R=\frac{2E_0+3E_0}{2} \] \[ \overline E_R=\frac52E_0 \] Thus, \[ \Phi_R=\frac52E_0\left(\frac{L^2}{2}\right) \] \[ \Phi_R=\frac54E_0L^2 \] Their sum agrees with the original total: \[ \Phi_L+\Phi_R=2E_0L^2 \] \(\textbf{Reverse only the left-half orientation:}\) Reversing a chosen area vector changes the sign of its flux. The right-half flux remains unchanged. Therefore, \[ \Phi_f=-\Phi_L+\Phi_R \] \[ \Phi_f=-\frac34E_0L^2+\frac54E_0L^2 \] \[ \Phi_f=\frac12E_0L^2 \] \(\textbf{Compare the signed fluxes:}\) \[ \frac{\Phi_f}{\Phi_i}=\frac{E_0L^2/2}{2E_0L^2} \] \[ \frac{\Phi_f}{\Phi_i}=\frac14 \] \(\textbf{Physical check:}\) The right half has greater positive flux because its field component is stronger. Reversing the weaker left contribution reduces the total but does not make it negative. \(\textbf{Final answer:}\) \[ \boxed{\frac{\Phi_f}{\Phi_i}=\frac14} \]