Physics MCQ Quiz - Electric Charges and Fields
1. How many protons are needed to make \(1\text{ nC}\) charge?

Correct Answer: (D) \(6.25 \times 10^{9}\)

Formula: \(Q = n \cdot e \Rightarrow n = \frac{Q}{e}\)
Given \(Q = 1\text{ nC} = 10^{-9}\text{ C}\), \(e = 1.6 \times 10^{-19}\text{ C}\).
\(n = \frac{10^{-9}}{1.6 \times 10^{-19}} = 6.25 \times 10^{9}\)

2. Two charges \(+q\) and \(-q\) are located at points \((a,0)\) and \((3a, 0)\) on the x-axis. The dipole moment of the system has magnitude:

Correct Answer: (B) \(2qa\) and is along \(-x\)-axis.

Separation distance \(d = 3a - a = 2a\). Magnitude \(p = q(2a) = 2qa\). Direction is from \(-q\) at \((3a,0)\) to \(+q\) at \((a,0)\), which points along the negative x-axis.

3. The magnitude of electric field at a certain distance from a dipole on its axis is \(E_1\) and at the same distance on the equatorial plane of the dipole is \(E_2\). Then \(\frac{E_1}{E_2}\) is:

Correct Answer: (B) 2

For a short dipole, \(E_1 = \frac{1}{4\pi\epsilon_0}\frac{2p}{r^3}\) and \(E_2 = \frac{1}{4\pi\epsilon_0}\frac{p}{r^3}\). Ratio \(\frac{E_1}{E_2} = 2\).

4. An electric dipole having dipole moment of \(4 \times 10^{-9}\text{ Cm}\) is placed in a uniform electric field of \(5 \times 10^4\text{ N/C}\). If the angle between the dipole moment and the electric field is \(30^\circ\), then the magnitude of potential energy of the dipole is:

Correct Answer: (C) \(1.73 \times 10^{-4}\text{ J}\)

\(U = -pE \cos\theta = -(4 \times 10^{-9})(5 \times 10^4)\cos(30^\circ) = -20 \times 10^{-5} \times \frac{\sqrt{3}}{2} \approx -1.73 \times 10^{-4}\text{ J}\).

5. The dimensional formula of dipole moment is:

Correct Answer: (D) \([\text{LTA}]\)

\([p] = [q][2a] = [\text{A}\cdot\text{T}][\text{L}] = [\text{LTA}]\).

6. An electric dipole of dipole moment \(\vec{p}\) is placed in a uniform electric field \(\vec{E}\). The torque exerted by the field on the dipole at that instant is:

Correct Answer: (B) perpendicular to both \(\vec{E}\) and \(\vec{p}\).

Torque \(\vec{\tau} = \vec{p} \times \vec{E}\). By properties of vector cross product, \(\vec{\tau}\) is perpendicular to both \(\vec{p}\) and \(\vec{E}\).

7. A conducting wire connects two charged metallic spheres A and B of radii \(r_1\) and \(r_2\) respectively. The ratio of electric fields, \((E_A/E_B)\) at the surfaces of spheres A and B will be:

Correct Answer: (B) \(\frac{r_2}{r_1}\)

Since potentials are equal (\(V_A = V_B\)), \(\frac{Q_A}{r_1} = \frac{Q_B}{r_2} \Rightarrow \frac{Q_A}{Q_B} = \frac{r_1}{r_2}\).
Ratio of fields: \(\frac{E_A}{E_B} = \left(\frac{Q_A}{Q_B}\right)\left(\frac{r_2^2}{r_1^2}\right) = \left(\frac{r_1}{r_2}\right)\left(\frac{r_2^2}{r_1^2}\right) = \frac{r_2}{r_1}\).

8. A particle of mass \(m\) and charge \(q\) starts from rest and moves in an electric field \(\vec{E} = E_0 \hat{i}\). After travelling a distance \(x\) in the field along x-axis, the kinetic energy of the particle will be:

Correct Answer: (B) \(qE_0x\)

Using Work-Energy Theorem: \(\text{K.E.} = W = F \cdot x = (q E_0) x\).

9. An electric dipole \(\vec{P} = 2qa \hat{i}\) is placed in x-y plane, centred at origin. Let \(E_1\) and \(E_2\) be the magnitudes of electric field at a point distant \(r (r \gg a)\) from its centre, along its axis and on its equatorial plane, respectively. Then the value of \(\left(\frac{E_1}{E_2}\right)\) is:

Correct Answer: (C) 2

For a short dipole, \(E_1 = E_{\text{axial}} = \frac{2p}{4\pi\epsilon_0 r^3}\) and \(E_2 = E_{\text{equatorial}} = \frac{p}{4\pi\epsilon_0 r^3}\). Ratio \(\frac{E_1}{E_2} = 2\).

10. Two small identical metallic balls having charges \(q\) and \(-2q\) are kept far at a separation \(r\). They are brought in contact and then separated at distance \(\frac{r}{2}\). Compared to the initial force \(F\), they will now:

Correct Answer: (B) repel with a force \(F/2\)

Initial force \(F = \frac{k |q(-2q)|}{r^2} = \frac{2kq^2}{r^2}\).
Final charge on each sphere after contact \(q' = \frac{q + (-2q)}{2} = -\frac{q}{2}\).
New force \(F' = \frac{k (-q/2)^2}{(r/2)^2} = \frac{kq^2}{r^2} = \frac{F}{2}\). Since charges are like, force is repulsive.

11. Two point charges \(-Q\) and \(Q\) are located at points \((d, 0)\) and \((0, d)\) respectively, in x-y plane. The electric field \(\vec{E}\) at the origin will be:

Correct Answer: (C) \(\frac{1}{4\pi\epsilon_0}\frac{Q}{d^2}(\hat{i}-\hat{j})\)

Field due to \(-Q\) at \((d,0)\) points towards it: \(\vec{E}_1 = \frac{kQ}{d^2}\hat{i}\).
Field due to \(+Q\) at \((0,d)\) points away from it: \(\vec{E}_2 = \frac{kQ}{d^2}(-\hat{j})\).
Net field \(\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{1}{4\pi\epsilon_0}\frac{Q}{d^2}(\hat{i}-\hat{j})\).

12. An electron is moving with velocity \(v_0\hat{i}\). If a uniform electric field \(\vec{E} = E_0\hat{j}\) is set up in the region, the electron will:

Correct Answer: (C) describe a parabolic path

Acceleration acts perpendicular to initial velocity (\(a_y = -\frac{eE_0}{m}\)), resulting in a parabolic trajectory \(y = -kx^2\).

13. Three point charges are kept at the vertices of triangle ABC (\(2q\) at A, \(-q\) at B, and \(-q\) at C). The net electric field due to this system at the midpoint M of base BC will be:

Correct Answer: (C) \(\frac{q}{2\pi\epsilon_0 l^2}\) pointing along AM

Fields from charges at B and C cancel out at midpoint M. Distance \(AM = l\). Field due to \(2q\) at A is \(E = \frac{1}{4\pi\epsilon_0}\frac{2q}{l^2} = \frac{q}{2\pi\epsilon_0 l^2}\) directed away from A (along AM).