According to Coulomb’s Law, \(F \propto \displaystyle\frac{1}{r^2}\). If distance becomes \(2r\), the force becomes \(\displaystyle\frac{1}{(2)^2} = \displaystyle\frac{1}{4}\) of the original.
Correct Answer: B

For the net force to be zero, the forces from \(+Q\) and \(-4Q\) must be in opposite directions and equal in magnitude. This occurs outside the charges, closer to the smaller magnitude charge (\(+Q\)). Therefore, the point must be at \(x < 0\).
Correct Answer: B

Charge is quantized: \(Q = ne\), where \(n\) is an integer. \(4.8 \times 10^{-19} \text{ C} / (1.6 \times 10^{-19} \text{ C}) = 3\). All other options yield non-integers.
Correct Answer: B

Total charge is conserved: \(Q_{total} = +Q + (-3Q) = -2Q\). When identical spheres touch, the charge divides equally: \(-2Q / 2 = -Q\) each.
Correct Answer: A

By Newton’s Third Law, the force exerted by A on B is equal and opposite to the force exerted by B on A. If the interaction force increases to \(20 \text{ N}\) due to doubling a charge, both spheres feel that same \(20 \text{ N}\) force.
Correct Answer: C

This is charging by induction. The negative rod repels electrons to the ground. Removing the ground leaves the sphere with a deficit of electrons (positive charge).
Correct Answer: B

Electrons are the mobile charge carriers in solids. To become negative, the rod must gain electrons from the fur.
Correct Answer: B

Both can be charged, but insulators have tightly bound electrons that cannot flow freely, whereas conductors have a “sea” of free electrons.
Correct Answer: C

Charging by contact is called conduction. Electrons flow from the sphere to the rod, leaving the sphere positive.
Correct Answer: B

In an insulator, charges can’t move across the object, but they can shift within atoms/molecules, creating a dipoles.
Correct Answer: C

\(\vec{F} = q\vec{E}\). Since an electron has a negative charge (\(q < 0\)), the force is in the opposite direction of the electric field.
Correct Answer: B

The field from \(+Q\) points away (toward \(-Q\)). The field from \(-Q\) points toward \(-Q\). Both vectors add together, pointing toward the negative charge.
Correct Answer: C

The field follows an inverse square law: \(E \propto \displaystyle\frac{1}{r^2}\). Tripling the distance reduces the field by a factor of \(3^2 = 9\).
Correct Answer: B

On a conductor’s surface, any parallel component of \(E\) would move charges. Thus, in equilibrium, \(E\) must be purely perpendicular.
Correct Answer: C

\(a = \frac{F}{m} = \frac{qE}{m}\).
\(a_p = \frac{eE}{m_p}\).
\(a_\alpha = \frac{2eE}{4m_p} = \frac{eE}{2m_p}\).
Ratio \(a_p / a_\alpha = 2\).
Correct Answer: B

For an infinite line of charge, the electric field is given by \( E = \displaystyle\frac{\lambda}{2\pi\epsilon_0 r} \). Thus, \( E \) is inversely proportional to the distance \( r \).
Correct Answer: A

By symmetry, for every small charge element \( dq \) on the ring, there is an identical element on the opposite side. Their field vectors at the center are equal in magnitude but opposite in direction, canceling out perfectly.
Correct Answer: D

The electric field of an infinite sheet is \( E = \displaystyle\frac{\sigma}{2\epsilon_0} \). This expression does not depend on the distance \( z \) from the sheet.
Correct Answer: D

Due to symmetry, components parallel to the diameter cancel out. The remaining components add up along the axis of symmetry pointing away from the positive charge distribution.
Correct Answer: B

Option A uses the position \( x \) directly. Option B uses a coordinate starting at the left end of the rod. Both correctly represent the distance squared from the origin to a charge element \( dq = \lambda dx \).
Correct Answer: C

The area vector \( \vec{A} \) for a surface in the yz-plane is parallel to the x-axis (\( \hat{i} \)). Since \( \vec{E} \) and \( \vec{A} \) are parallel, \( \Phi = E_0 A \cos(0^\circ) = E_0 A \).
Correct Answer: B

For any closed surface in a uniform field with no enclosed charge, the number of field lines entering equals the number of field lines leaving. Thus, the net flux is zero.
Correct Answer: D

By Gauss’s Law, \( \Phi = q_{enc} / \epsilon_0 \). Since the enclosed charge hasn’t changed, the total flux remains the same regardless of the surface size.
Correct Answer: D

By convention, the infinitesimal area vector \( d\vec{A} \) for any closed surface always points radially outward and perpendicular to the surface.
Correct Answer: C

Flux is \( EA\cos\theta \), where \( \theta \) is the angle between \( \vec{E} \) and the normal (area vector). If the angle with the surface is \( 30^\circ \), then \( \theta = 90^\circ – 30^\circ = 60^\circ \). Note that \( \cos(60^\circ) = \sin(30^\circ) \).
Correct Answer: B

In a conductor in electrostatic equilibrium, all excess charge resides on the surface. Therefore, the enclosed charge for any Gaussian surface inside (\( r < R \)) is zero, and \( E = 0 \).
Correct Answer: D

Using Gauss’s Law: \( E(4\pi r^2) = \displaystyle\frac{\rho(\displaystyle\frac{4}{3}\pi r^3)}{\epsilon_0} \). Simplifying gives \( E = \displaystyle\frac{\rho r}{3\epsilon_0} \). Thus, \( E \) is directly proportional to \( r \).
Correct Answer: B

Applying Gauss’s Law to a cylinder of length \( L \): \( E(2\pi r L) = \displaystyle\frac{\sigma (2\pi R L)}{\epsilon_0} \). Solving for \( E \) gives \( E = \displaystyle\frac{\sigma R}{\epsilon_0 r} \).
Correct Answer: A

For the electric field inside the conductor material to be zero, a Gaussian surface within the shell must enclose zero net charge. Since there is \( -Q \) at the center, the inner surface must have \( +Q \).
Correct Answer: B

Both a cylinder and a cube (with faces parallel/perpendicular to the field) work effectively because they maintain a constant field magnitude over the end caps and have no flux through the sides.
Correct Answer: D