Exams › JEE Main › Physics

Two points P and Q are maintained at the potentials of 10 V and −4 V, respectively. The work done in moving 100 electrons from P to Q is:

1. 9.60 × 10−17 J
2. −2.24 × 10−16 J
3. 2.24 × 10−16 J
4. −9.60 × 10−17 J

Correct answer: 2.24 × 10−16 J

Solution

Charge moved q = 100 x (-1.6e-19) = -1.6e-17 C. W = q(V_Q - V_P) = -1.6e-17 x (-4 - 10) = -1.6e-17 x (-14) = +2.24e-16 J.

Related JEE Main Physics questions

• If an electric field vector E exists in a material whose dielectric constant is K, and ε0 denotes the permittivity of free space, which expression gives the electric displacement vector?
• If a unit positive charge is taken from point A to point B along the same equipotential surface, then what can be said about the potential difference V_A - V_B?
• Two thin metallic spherical shells are placed concentrically with radii R1 and R2, where R1 > R2. If the shells carry charges Q1 and Q2 respectively, what is the electric potential at a point located at distance r such that R2 < r < R1? (Take k = 1/(4πϵ0))
• An alpha particle is made to move through a potential difference of 10⁶ V. What kinetic energy does it gain?
• A parallel-plate capacitor has its entire gap filled with a dielectric whose dielectric constant changes with position x according to K(x) = K0 + λx, where λ is a constant. How is its capacitance C related to the vacuum capacitance C0?
• Two thin circular rings of radius R each are kept on the same axis with their centres separated by a distance d. One ring carries charge +q and the other carries charge −q. What is the potential difference between the centres of the two rings?

⚔️ Practice JEE Main Physics free + battle 1v1 →