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

Work done will be
`14.7* 10^(-6)J` and it will be positive

Step-by-step explanation:

We have given charge
`2.1* 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* 10^(-6)* 7=14.7* 10^(-6)J`

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