Electric Charges and Fields

Answer explanation

The force between two charges is given by Coulomb's law: F = k * (q1 * q2) / r^2. If charges are doubled (2q1, 2q2) and distance is halved (r/2), the new force becomes F' = k * (4q1 * q2) / (r/2)^2 = 16F, thus 16\phi.

An electric dipole is placed in a uniform field. The resultant force acting on it -------.

• always be zero

• depends on its relative position

• never be zero

• depends on its dipole moment

Answer explanation

In a uniform electric field, the forces on the positive and negative charges of the dipole are equal and opposite, resulting in a net force of zero. Therefore, the correct answer is that the resultant force acting on it always be zero.

An electric dipole is placed at an angle of 30° with an electric field intensity 2 x 10⁵ N/C. It experiences a torque equal to 4 N-m. The charge on the dipole, if the dipole length is 2 cm, is -------.

• 8 mC

• 2 mC

• 5 mC

Answer explanation

The torque (τ) on a dipole is given by τ = pE sin(θ). Here, p = qL, where q is the charge and L is the length. Rearranging gives q = τ/(EL sin(θ)). Substituting τ = 4 N-m, E = 2 x 10^5 N/C, L = 0.02 m, and θ = 30°, we find q = 2 mC.

Answer explanation

Using Gauss's law, the electric flux \( \Phi_E = \frac{Q}{\epsilon_0} \). Here, \( Q = 10 \mu C = 10 \times 10^{-6} C \) and \( \epsilon_0 \approx 8.85 \times 10^{-12} \frac{C^2}{Nm^2} \). Calculating gives \( \Phi_E \approx 1.88 \times 10^5 \frac{Nm^2}{C} \), matching the correct choice.

Charge q₂ of mass m revolves around a stationary charge q₁ in a circular orbit of radius r. The orbital periodic time of q₂ would be -------.

Answer explanation

To find the orbital period of charge q2, we use the centripetal force due to electrostatic attraction. The formula derived leads to the periodic time T = \sqrt{\frac{4\pi^2mr^3}{kq_1q_2}}, confirming the correct choice.

Electric field produced due to an infinitely long straight uniformly charged wire at perpendicular distance of 2 cm is 3 x 10⁸ N/C. Then linear charge density on the wire is -------. (k = 9 x 10⁹ SI)

Answer explanation

The electric field E due to a line charge is given by E = (k * λ) / r. Rearranging gives λ = (E * r) / k. Substituting E = 3 x 10^8 N/C, r = 0.02 m, and k = 9 x 10^9 SI, we find λ = 333 μC/m, which is the correct answer.

Consider the electric field lines shown in the diagram below. From the diagram, it is apparent that object A is ____ and object B is ____.

• +,+

• +,-

• -,+

• -,-

Answer explanation

Object A has electric field lines pointing towards it, indicating it is negatively charged (-). Object B has lines pointing away, indicating it is positively charged (+). Thus, the correct answer is -, +.

Based on the arrangement of the electric field lines in the diagram, what are the charges?

• both charges are the same

• the charges are opposite

• they are neutral; no charge

Answer explanation

The arrangement of electric field lines indicates that both charges are the same. If they were opposite, the lines would converge, while parallel lines suggest like charges, confirming that both charges are indeed the same.

A moving electron is deflected by two oppositely charged parallel plates, as shown in the diagram above.

The electric field between the plates is directed from

• C to D

• D to C

• A to B

• B to A

Answer explanation

The electric field between the plates is directed from the positive plate (C) to the negative plate (D). Therefore, the correct answer is C to D.

What is the magnitude of the electric field intensity at a point where a proton experiences an electrostatic force of magnitude 2.30 x 10^-25 newton?

• 1.44 x 10^-6 N/C

• 3.68 x 10⁶ N/C

• 1.44 x 10⁴⁴ N/C

• 3.68 x 10^-44 N/C

Answer explanation

The electric field intensity (E) can be calculated using the formula F = qE, where F is the force and q is the charge of the proton (1.6 x 10^-19 C). Rearranging gives E = F/q. Substituting F = 2.30 x 10^-25 N, we find E = 1.44 x 10^-6 N/C.