Class 12 Physics MCQs | Chapter 1: Electric Charges and Fields – Part 5

Explanation: If distance is doubled and the field becomes \(\frac{1}{8}\), the field varies as \(\frac{1}{r^3}\). A far dipole field follows \(E\propto r^{-3}\). A point charge would become \(\frac{1}{4}\) when distance is doubled. An infinite line charge would become \(\frac{1}{2}\), and an infinite plane sheet would remain unchanged. The factor change under doubling distance is a quick way to identify the power law. The common mistake is to choose point charge whenever field decreases, but the decrease factor here is inverse-cube.

Explanation: If a field becomes half when distance is doubled, the field follows \(E\propto r^{-1}\). This is the distance dependence for an infinitely long uniformly charged line: \(E=\frac{\lambda}{2\pi\varepsilon_0 r}\). A point charge would give one-fourth on doubling distance. A far dipole would give one-eighth. An infinite plane sheet would show no change with distance. The doubling-distance test is a useful graph and proportionality shortcut.

Explanation: For a point charge, \(E\propto r^{-2}\), so doubling \(r\) makes \(E\) become \(\frac{E}{4}\); row P is correct. For an infinite line charge, \(E\propto r^{-1}\), so doubling \(r\) gives \(\frac{E}{2}\); row Q is correct. For an infinite plane sheet, \(E\) is independent of \(r\), so row R is correct. For a far dipole, \(E\propto r^{-3}\), so doubling \(r\) gives \(\frac{E}{8}\); row S is correct. This table summarizes the main electrostatic proportionality patterns. The safest method is to apply the power of \(r\) directly.

Explanation: For points outside a spherically symmetric charged distribution, Gauss's law gives the same field as if the total charge were concentrated at the centre. This applies to both a uniformly charged thin spherical shell and a uniformly charged solid sphere. Since both have the same total charge \(Q\) and the same outside observation distance \(r\), the field magnitudes are equal. Each is \(E=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\). The difference between shell and solid sphere appears inside the radius \(R\), not outside. The region check \(r\gt R\) is the deciding condition.

Explanation: At \(r=\frac{R}{2}\), the point lies inside both spherical distributions. For a thin spherical shell, all charge is on the surface at \(r=R\), so a Gaussian surface at \(r=\frac{R}{2}\) encloses no charge and the field is zero. For a uniformly charged solid sphere, charge is spread throughout the volume, so the smaller Gaussian sphere encloses some charge. The inside solid-sphere field varies as \(E=\frac{1}{4\pi\varepsilon_0}\frac{Qr}{R^3}\), which is non-zero for \(r=\frac{R}{2}\). The key difference is surface charge distribution versus volume charge distribution. The common mistake is to apply the shell zero-field result to every spherical object.

Explanation: \( \textbf{Electric force:} \) The magnitude of force on \(+q\) due to \(+Q\) is \[ F=k\frac{Qq}{r^2} \] \( \textbf{Newton's second law:} \) \[ F=ma \] \( \textbf{Acceleration relation:} \) \[ a=\frac{F}{m} \] \( \textbf{Substitute electric force:} \) \[ a=\frac{1}{m}\left(k\frac{Qq}{r^2}\right) \] \( \textbf{Result:} \) \[ a=\frac{kQq}{mr^2} \] \( \textbf{Direction note:} \) Since both charges are positive, the acceleration is away from the fixed charge \(+Q\). \( \textbf{Final answer:} \) The initial acceleration magnitude is \(\frac{kQq}{mr^2}\).

Explanation: \( \textbf{Given:} \) \(m=2.0\times10^{-6}\,\text{kg}\), \(q=3.0\,\mu\text{C}=3.0\times10^{-6}\,\text{C}\), and \(E=4.0\times10^4\,\text{N C}^{-1}\). \( \textbf{Required:} \) Acceleration magnitude \(a\). \( \textbf{Electric force:} \) \[ F=qE \] \( \textbf{Substitution for force:} \) \[ F=(3.0\times10^{-6})(4.0\times10^4) \] \( \textbf{Force calculation:} \) \[ F=12.0\times10^{-2}\,\text{N}=0.12\,\text{N} \] \( \textbf{Newton's second law:} \) \[ a=\frac{F}{m} \] \( \textbf{Substitution:} \) \[ a=\frac{0.12}{2.0\times10^{-6}} \] \( \textbf{Final calculation:} \) \[ a=6.0\times10^4\,\text{m s}^{-2} \] \( \textbf{Final answer:} \) The acceleration magnitude is \(6.0\times10^4\,\text{m s}^{-2}\).

Explanation: \( \textbf{Given:} \) \(\Phi_E=4.0\times10^3\,\text{N m}^2\text{C}^{-1}\), and \(\varepsilon_0=8.85\times10^{-12}\,\text{C}^2\text{N}^{-1}\text{m}^{-2}\). \( \textbf{Required:} \) Enclosed charge \(q_{\text{enc}}\). \( \textbf{Gauss's law:} \) \[ \Phi_E=\frac{q_{\text{enc}}}{\varepsilon_0} \] \( \textbf{Rearranged form:} \) \[ q_{\text{enc}}=\varepsilon_0\Phi_E \] \( \textbf{Substitution:} \) \[ q_{\text{enc}}=(8.85\times10^{-12})(4.0\times10^3) \] \( \textbf{Coefficient multiplication:} \) \[ 8.85\times4.0=35.4 \] \( \textbf{Power multiplication:} \) \[ 10^{-12}\times10^3=10^{-9} \] \( \textbf{Final simplification:} \) \[ q_{\text{enc}}=35.4\times10^{-9}\,\text{C}=3.54\times10^{-8}\,\text{C} \] \( \textbf{Final answer:} \) The enclosed charge is \(3.54\times10^{-8}\,\text{C}\).

Explanation: \( \textbf{Given:} \) \(q_1=4.0\times10^{-6}\,\text{C}\), \(q_2=6.0\times10^{-6}\,\text{C}\), \(r=0.30\,\text{m}\), \(K=3\), and \(m=2.0\times10^{-4}\,\text{kg}\). \( \textbf{Required:} \) Acceleration of the body carrying \(+4.0\,\mu\text{C}\). \( \textbf{Force in a medium:} \) \[ F=\frac{k|q_1q_2|}{Kr^2} \] \( \textbf{Why this formula applies:} \) The charges are point charges in a dielectric medium, so the vacuum force is divided by \(K\). \( \textbf{Charge product:} \) \[ |q_1q_2|=(4.0\times10^{-6})(6.0\times10^{-6})=24.0\times10^{-12} \] \( \textbf{Distance and medium factor:} \) \[ Kr^2=3(0.30)^2=3(0.090)=0.270 \] \( \textbf{Substitution:} \) \[ F=\frac{(9.0\times10^9)(24.0\times10^{-12})}{0.270} \] \( \textbf{Force calculation:} \) \[ F=\frac{0.216}{0.270}=0.80\,\text{N} \] \( \textbf{Acceleration relation:} \) \[ a=\frac{F}{m} \] \( \textbf{Final calculation:} \) \[ a=\frac{0.80}{2.0\times10^{-4}}=4.0\times10^3\,\text{m s}^{-2} \] \( \textbf{Final answer:} \) The acceleration magnitude is \(4.0\times10^3\,\text{m s}^{-2}\).

Explanation: \( \textbf{Given:} \) Fixed charge \(Q=+3.0\times10^{-6}\,\text{C}\), moving charge \(q=-2.0\times10^{-6}\,\text{C}\), \(r=0.30\,\text{m}\), and \(m=5.0\times10^{-6}\,\text{kg}\). \( \textbf{Required:} \) Initial acceleration of the negative charge. \( \textbf{Electric force magnitude:} \) \[ F=k\frac{|Qq|}{r^2} \] \( \textbf{Substitution:} \) \[ F=\frac{(9.0\times10^9)(3.0\times10^{-6})(2.0\times10^{-6})}{(0.30)^2} \] \( \textbf{Simplification:} \) \[ F=\frac{(9.0\times10^9)(6.0\times10^{-12})}{0.090} \] \[ F=\frac{0.054}{0.090}=0.60\,\text{N} \] \( \textbf{Acceleration:} \) \[ a=\frac{F}{m} \] \( \textbf{Substitution:} \) \[ a=\frac{0.60}{5.0\times10^{-6}}=1.2\times10^5\,\text{m s}^{-2} \] \( \textbf{Direction check:} \) The charges are unlike, so the negative charge is attracted toward the fixed positive charge. \( \textbf{Position check:} \) Since the negative charge is to the right of the positive charge, attraction is leftward. \( \textbf{Final answer:} \) The initial acceleration is \(1.2\times10^5\,\text{m s}^{-2}\) leftward.

Explanation: \( \textbf{Gauss-law principle:} \) Net flux through a closed surface depends only on algebraic enclosed charge. \( \textbf{Enclosed charges:} \) \(+5.0\,\mu\text{C}\), \(-8.0\,\mu\text{C}\), and \(+1.0\,\mu\text{C}\). \( \textbf{External charge:} \) The outside charge \(+10.0\,\mu\text{C}\) may affect local \(\vec{E}\), but it does not enter \(q_{\text{enc}}\). \( \textbf{Algebraic enclosed charge:} \) \[ q_{\text{enc}}=(+5.0-8.0+1.0)\,\mu\text{C} \] \( \textbf{Simplification:} \) \[ q_{\text{enc}}=-2.0\,\mu\text{C} \] \( \textbf{Gauss's law:} \) \[ \Phi_E=\frac{q_{\text{enc}}}{\varepsilon_0} \] \( \textbf{Substitution:} \) \[ \Phi_E=\frac{-2.0\,\mu\text{C}}{\varepsilon_0} \] \( \textbf{Sign check:} \) The negative sign means net inward flux is greater than net outward flux. \( \textbf{Final answer:} \) The net flux is \(\frac{-2.0\,\mu\text{C}}{\varepsilon_0}\).

Explanation: \( \textbf{Given:} \) \(\lambda=5.0\times10^{-6}\,\text{C m}^{-1}\), \(q=2.0\times10^{-6}\,\text{C}\), \(r=0.25\,\text{m}\), and \(\frac{1}{2\pi\varepsilon_0}=1.8\times10^{10}\,\text{N m}^2\text{C}^{-2}\). \( \textbf{Required:} \) Force magnitude on the charge. \( \textbf{Field of infinite line charge:} \) \[ E=\left(\frac{1}{2\pi\varepsilon_0}\right)\frac{\lambda}{r} \] \( \textbf{Substitution for field:} \) \[ E=(1.8\times10^{10})\frac{5.0\times10^{-6}}{0.25} \] \( \textbf{Simplify charge density over distance:} \) \[ \frac{5.0\times10^{-6}}{0.25}=2.0\times10^{-5} \] \( \textbf{Field calculation:} \) \[ E=(1.8\times10^{10})(2.0\times10^{-5})=3.6\times10^5\,\text{N C}^{-1} \] \( \textbf{Force relation:} \) \[ F=qE \] \( \textbf{Substitution for force:} \) \[ F=(2.0\times10^{-6})(3.6\times10^5)=0.72\,\text{N} \] \( \textbf{Final answer:} \) The force magnitude is \(0.72\,\text{N}\).

Explanation: \( \textbf{Given:} \) \(q=3.0\,\mu\text{C}=3.0\times10^{-6}\,\text{C}\), \(\sigma=6.0\times10^{-8}\,\text{C m}^{-2}\), and \(\varepsilon_0=8.0\times10^{-12}\,\text{C}^2\text{N}^{-1}\text{m}^{-2}\). \( \textbf{Required:} \) Force magnitude on the charge. \( \textbf{Field due to one infinite sheet:} \) \[ E=\frac{\sigma}{2\varepsilon_0} \] \( \textbf{Substitution for field:} \) \[ E=\frac{6.0\times10^{-8}}{2(8.0\times10^{-12})} \] \( \textbf{Denominator:} \) \[ 2(8.0\times10^{-12})=1.6\times10^{-11} \] \( \textbf{Field magnitude:} \) \[ E=\frac{6.0\times10^{-8}}{1.6\times10^{-11}}=3.75\times10^3\,\text{N C}^{-1} \] \( \textbf{Force relation:} \) \[ F=qE \] \( \textbf{Substitution:} \) \[ F=(3.0\times10^{-6})(3.75\times10^3) \] \( \textbf{Final calculation:} \) \[ F=11.25\times10^{-3}\,\text{N}=1.125\times10^{-2}\,\text{N} \] \( \textbf{Final answer:} \) The force magnitude is \(1.125\times10^{-2}\,\text{N}\).

Explanation: \( \textbf{Inside-field formula:} \) For a uniformly charged solid sphere, \[ E=\frac{\rho r}{3\varepsilon_0} \] \( \textbf{Force on charge:} \) \[ F=qE \] \( \textbf{Substitute the field:} \) \[ F=q\left(\frac{\rho r}{3\varepsilon_0}\right) \] \( \textbf{Simplified force:} \) \[ F=\frac{q\rho r}{3\varepsilon_0} \] \( \textbf{Newton's second law:} \) \[ a=\frac{F}{m} \] \( \textbf{Substitution:} \) \[ a=\frac{q\rho r}{3m\varepsilon_0} \] \( \textbf{Dependence check:} \) The acceleration is proportional to \(r\) inside the uniformly charged solid sphere. \( \textbf{Final answer:} \) The acceleration magnitude is \(\frac{q\rho r}{3m\varepsilon_0}\).

Explanation: \( \textbf{Given:} \) \(q=4.0\times10^{-6}\,\text{C}\), separation \(d=5.0\,\text{cm}=5.0\times10^{-2}\,\text{m}\), \(E=2.0\times10^5\,\text{N C}^{-1}\), and \(\theta=60^\circ\). \( \textbf{Required:} \) Torque magnitude \(\tau\). \( \textbf{Dipole moment:} \) \[ p=qd \] \( \textbf{Substitution for} \) \(p\): \[ p=(4.0\times10^{-6})(5.0\times10^{-2})=2.0\times10^{-7}\,\text{C m} \] \( \textbf{Torque formula:} \) \[ \tau=pE\sin\theta \] \( \textbf{Use} \) \(\sin60^\circ\approx0.866\): \[ \tau=(2.0\times10^{-7})(2.0\times10^5)(0.866) \] \( \textbf{Intermediate product:} \) \[ (2.0\times10^{-7})(2.0\times10^5)=4.0\times10^{-2} \] \( \textbf{Final calculation:} \) \[ \tau=(4.0\times10^{-2})(0.866)=3.46\times10^{-2}\,\text{N m} \] \( \textbf{Final answer:} \) The torque magnitude is \(3.46\times10^{-2}\,\text{N m}\).

Explanation: \( \textbf{Given fields:} \) \(E_x=6.0\times10^4\,\text{N C}^{-1}\), and \(E_y=8.0\times10^4\,\text{N C}^{-1}\). \( \textbf{Net field magnitude:} \) Since the fields are perpendicular, \[ E_{\text{net}}=\sqrt{E_x^2+E_y^2} \] \( \textbf{Substitution:} \) \[ E_{\text{net}}=\sqrt{(6.0\times10^4)^2+(8.0\times10^4)^2} \] \( \textbf{Simplification:} \) \[ E_{\text{net}}=10.0\times10^4\,\text{N C}^{-1}=1.0\times10^5\,\text{N C}^{-1} \] \( \textbf{Charge magnitude:} \) \[ |q|=2.0\,\mu\text{C}=2.0\times10^{-6}\,\text{C} \] \( \textbf{Force magnitude:} \) \[ F=|q|E_{\text{net}} \] \( \textbf{Substitution:} \) \[ F=(2.0\times10^{-6})(1.0\times10^5) \] \( \textbf{Final calculation:} \) \[ F=2.0\times10^{-1}\,\text{N}=0.20\,\text{N} \] \( \textbf{Direction note:} \) Since the charge is negative, the force direction is opposite to the net electric field. \( \textbf{Final answer:} \) The force magnitude is \(0.20\,\text{N}\).

Explanation: Row P is correct because a charge \(q\) in an electric field experiences force \(\vec{F}=q\vec{E}\). Row Q is correct because Gauss's law gives net flux through a closed surface as \(\Phi_E=\frac{q_{\text{enc}}}{\varepsilon_0}\). Row R is correct because torque on a dipole in a uniform field is \(\vec{\tau}=\vec{p}\times\vec{E}\). Row S is correct because the electric field inside a conductor in electrostatic equilibrium is zero. These relations belong to different parts of electrostatics but are often combined in mixed problems. The key step is choosing the relation from the physical situation, not from symbol similarity.

Explanation: If electric field becomes \(\frac{1}{4}\) when distance is doubled, the field follows \(E\propto r^{-2}\). This is the distance dependence of a point charge. If electric field remains unchanged when distance is doubled, the field is independent of distance. This is the ideal behaviour of a single infinite plane sheet. An infinite line charge would give \(\frac{1}{2}\) on doubling distance, and a far dipole would give \(\frac{1}{8}\). The doubling-distance test is a quick way to identify the field law. The two graphs therefore match a point charge and an infinite plane sheet.

Explanation: In electrostatic equilibrium, the electric field inside the conducting material must be zero. The off-centre charge in the cavity can induce charge on the inner and outer surfaces of the conductor. These induced charges arrange themselves so that the field inside the metal itself is cancelled. The field inside the cavity need not be zero because the point charge is present there. Excess or induced charge resides on surfaces, not throughout the conducting material. The wording mistake is distinguishing the cavity region from the conducting material.

Explanation: \( \textbf{Initial force:} \) \[ F=k\frac{|q_1q_2|}{r^2} \] \( \textbf{Medium effect:} \) A dielectric constant \(K=4\) divides the force by \(4\). \( \textbf{Distance effect:} \) The new separation is \(\frac{r}{3}\), so the inverse-square factor increases by \(9\). \( \textbf{New force expression:} \) \[ F'=\frac{k|q_1q_2|}{4(r/3)^2} \] \( \textbf{Square the new distance:} \) \[ (r/3)^2=\frac{r^2}{9} \] \( \textbf{Simplification:} \) \[ F'=\frac{k|q_1q_2|}{4r^2/9} \] \[ F'=\frac{9}{4}k\frac{|q_1q_2|}{r^2} \] \( \textbf{Comparison:} \) \[ F'=\frac{9F}{4} \] \( \textbf{Final answer:} \) The new force is \(\frac{9F}{4}\).