Question

Find the voltage change when (a) an electric field does 12 J of work on a 0.0001-C charge and (b) the same electric field does 24 J of work on a 0.0002-C charge.

Answer 1

Answer:

a. $Change in Voltage=12*10^{4}V$

b. $Change in Voltage=12*10^{4}V$

Explanation:

The work done in moving an electric charge round a circuit is express as

$workdone=voltage*charge \\Wd=v*q$

The voltage is in-turn define as the electric potential energy per unit charge.

$Voltage=\frac{potitntial energy }{charge}$

a. for a 12J work done  on a charge of value 0.0001C, we can compute the voltage change as

$Voltage=\frac{potitntial energy }{charge}\\Voltage=\frac{12J}{0.0001C}\\ Voltage=12,0000J/C\\ Change in Voltage=12*10^{4}V$

a. for a 24J work done  on a charge of value 0.0002C, we can compute the voltage change as

$Voltage=\frac{potitntial energy }{charge}\\Voltage=\frac{24J}{0.0002C}\\ Voltage=12,0000J/C\\Change in Voltage=12*10^{4}V$