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Question: A ball of mass 1 gm carrying a charge $10^{-8}$ C moves only under the influence of an electrostatic...

A ball of mass 1 gm carrying a charge 10−810^{-8} C moves only under the influence of an electrostatic field from a point A at potential 700 V to a point B at zero potential. The change in its kinetic energy is

A

−7×10−6-7\times 10^{-6} erg

B

−7×10−6-7\times 10^{-6} J

C

7×10−67\times 10^{-6} J

D

7×10−67\times 10^{-6} erg

Answer

7×10−67\times 10^{-6} J

Explanation

Solution

The change in kinetic energy of a charge moving in an electrostatic field is equal to the work done by the electrostatic field. The work done is given by W=q(VA−VB)W = q(V_A - V_B), where qq is the charge, VAV_A is the initial potential, and VBV_B is the final potential.

Given q=10−8q = 10^{-8} C, VA=700V_A = 700 V, VB=0V_B = 0 V.

Change in kinetic energy ΔKE=q(VA−VB)=(10−8 C)(700 V−0 V)=700×10−8 J=7×10−6 J\Delta KE = q(V_A - V_B) = (10^{-8} \text{ C})(700 \text{ V} - 0 \text{ V}) = 700 \times 10^{-8} \text{ J} = 7 \times 10^{-6} \text{ J}.