An electric field of 20 N C^ -1 is present along the x -axis in space. Calculate the potential difference V_B - V_A when the points A and B are given by: (a) A = (0, 0) ; B = (4 m, 2 m) (b) A = (4 m, 2 m) ; B = (6 m, 5

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Correct answer: D. (a) -80 V , (b) -40 V , (c) -120 V

The relation between potential difference and electric field is given by

V_B - V_A = -\int_A^{B} \vec{E} \cdot d\vec{r}

.

Since the electric field is uniform and directed along the

x

-axis,

\vec{E} = 20 \hat{i} \text{ N C}^{-1}

.

The potential difference becomes

V_B - V_A = -E_x (x_B - x_A) = -20 (x_B - x_A)

.

For case (a):

x_A = 0 \text{ m}

,

x_B = 4 \text{ m}

V_B - V_A = -20(4 - 0) = -80 \text{ V}

For case (b):

x_A = 4 \text{ m}

,

x_B = 6 \text{ m}

V_B - V_A = -20(6 - 4) = -40 \text{ V}

For case (c):

x_A = 0 \text{ m}

,

x_B = 6 \text{ m}

V_B - V_A = -20(6 - 0) = -120 \text{ V}

An electric field of 20 N C^ -1 is present along the x -axis in...

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