Question

A particle of mass$1g$ and electric charge ${10}^{-8}C$ travels from a point $A$ having electric potential $0V$ to point $B$ having $600V$ electrical potential . What would be the change in its kinetic energy?

• A. $-6\times {10}^{-6}erg$
• B. $-6\times {10}^{-6}J$
• C. $6\times {10}^{-6}erg$
• D. $6\times {10}^{-6}J$

Answer 1

The correct option is D $6\times {10}^{-6}J$

Kinetic Energy $=q\Delta V$ , where $\Delta V=V_{final}-V_i$

KE $={10}^{-8}\times (600-0)$

KE $=6\times {10}^{-6}J$