Question

Calculate the work done in moving a charge of 2 micro coulomb from point A to B in an electric field where potential difference of A is 10 volts and potential difference of B is 10^6+10 volts.

Answer 1

Given

charge = 2 C

Potential difference V (A) = 10 V

potential difference V (B) = 5 V

V(A) - V(B) = W/Q

⇒10 - (-5) = W/2

⇒W = 30 J

Work done here = 30 J

Answer 2

2 J.

Explanation:

Given the moving charge is, $q = 2\mu C =2\times10^-^6C$.

The potential at point A, V = 10 volts.

The potential at point B, v = 1000010 volts.

The relationship between work done,potential difference and charge is

$W=q\Delta V$

substitute the given values, we get

$W=2\times10^-^6C(1000010V-10V)$

$W=2\times10^-^6C \times1000000V$

W = 2 J.

Thus, the work done in moving a charge of 2 micro coulomb from point A to B is 2J.

#Learn More: potential difference, work done.

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