Question

Consider the situation of the previous problem. A charge of −2.0 × 10−4 C is moved from point A to point B. Find the change in electrical potential energy UB − UA for the cases (a), (b) and (c).

Answer 1

The change in the electrical potential energy $(U_B-U_A)$ for the cases listed below :

Case (A) : $(U_B-U_A)$ = 0.016 J.

Case (B) : $(U_B-U_A)$ = 0.008 J.

Case (C) : $(U_B-U_A)$ = 0.024 J.

Explanation:

Given that ,            

                $q=-2.0 \times 10^{-4} \mathrm{C}$

                "q" is magnitude of a charge

Case (A) :

               The electrical field direction "x".

               Potential difference of (0, 0) to (4,2)

               dV = -E.dx = -20 × 4 = -80 V

Potential energy $\left(U_{B}-U_{A}\right)$ for points A to B = dv × q

              $\left(U_{B}-U_{A}\right)=(-80) \times\left(-2.0 \times 10^{-4} C\right)$

                 $U_{B}-U_{A}=160 \times 10^{-4}=0.016 J$

Case (B)  :

               A = ( 4m,2m)

               B = (6m, 5m)

               dv = -E.dx = -20×2 = -40 v

Potential energy $\left(U_{B}-U_{A}\right)$ for points A to B = dv × q

               $\left(U_{B}-U_{A}\right)=(-40) \times\left(-2.0 \times 10^{-4} C\right)$

                  $U_{B}-U_{A}=80 \times 10^{-4}=0.008 J$

Case (c)  :

               A = ( 0m,0m)

               B = (6m, 5m)

               dv = -E.dx = -20×6 = -120 v

Potential energy $\left(U_{B}-U_{A}\right)$ for points A to B = dv × q

            $\left(U_{B}-U_{A}\right)=(-120) \times\left(-2.0 \times 10^{-4} C\right)$

               $U_{B}-U_{A}=240 \times 10^{-4}=0.024 J$

Thus, the electrical potential energy $(U_B - U_A)$ for the cases (A), (B) and (C) has been determined.