Question

Find the work done "by" the electric field on a positively charged point particle with a charge of 2.1x10^-6 C as it is moved from a potential of 15.0 V to one of 8.0 V. (Include the sign of the value in your answer.)

Answer 1

Answer

Work done will be $14.7\times 10^{-6}J$ and it will be positive

Explanation:

We have given charge $2.1\times 10^{-6}C$

We have to find work done in moving the charge from 15 volt to 8 volt

Let $V_1=15V\ and\ V_2=8volt$

So potential difference $V=V_1-V_2=15-8=7volt$

We know that work done $W=QV$, here Q is charge and V is potential difference

So work done $W=QV=2.1\times 10^{-6}\times 7=14.7\times 10^{-6}J$

It will be positive work done because work is done in moving charge from higher potential to lower potential