Question

Points A and B lie within a region of space where there is a uniform electric field that has no x- or z-component; only the y-component Ey is non zero. Point A is at y=8.00 cm and point B is at y=15.0 cm. The potential difference between B and A is VB−VA=+12.0 V, so point B is at higher potential than point A. (a) Is Ey positive or negative? (b) What is the magnitude of the electric field? (c) Point C has coordinates x=5.00 cm, y=5.00 cm. What is the potential difference between points B and C?

Answer 1

Answer:

(a) Ey is negative

(b) The magnitude of the electric field is E = 171.429 V/m

(c) The potential difference between points B and C is 17.1429 V

Explanation:

(a) Here, we have the potentials given by;

$V_A - V_B = +12.0V$ with point A at y = 8.00 cm and point B at point y = 15.0 cm

where point B is at a higher potential than point A, that is the electric potential is from;

B with y = 15.0 cm to A with y = 8.0 cm which means

$E_y$ decreases as y increases or $E_y$  is negative.

(b) The magnitude of the electric field is given by

The work done to move a charge from B to A is

$W_{BA} = - \Delta U$ where

$\Delta U = U_a -U_b = q_0E(y_b-y_a)$

$V_{BA} = \frac{\Delta U}{q_0} = \frac{q_0E(y_b-y_a)}{q_0} = E(y_b-y_a)$

∴ $E = \frac{V_{BA}}{(y_b-y_a)}$

$E = \frac{12 \hspace{0.09cm}V}{(0.015\hspace{0.09cm} m -0.008\hspace{0.09cm} m)}$

E = 171.429 V/m

(c) Here we have point C x = 5.00 cm and y = 5.00 cm

Therefore we have the distance from B to C given by

$y_b-y_c = 15.00 \hspace{0.09cm}cm - 5.00 \hspace{0.09cm}cm = 10.00 \hspace{0.09cm} cm$

Where 10.00 cm = 0.01 m

E = V/Δy

Therefore, V = Δy·E

For $V_{BC}$, Δy = $y_b-y_c = 0.01 \hspace{0.09cm} m$ and we have,

$V_{BC} = E\times (y_b-y_c)$

$V_{BC} = 171.429\times (0.015-0.005) = 17.1429\hspace{0.09cm}V$