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(B) The potential difference between points $B$ and $C$ is $\Delta V = V_C - V_B = 20 \, V - 15 \, V = 5 \, V$.Since the electric field $\vec{E}$ is u

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(B) The potential difference between points $B$ and $C$ is $\Delta V = V_C - V_B = 20 \, V - 15 \, V = 5 \, V$.Since the electric field $\vec{E}$ is uniform and in the plane of the triangle,the potential difference is given by $\Delta V = E \cdot d \cos 	heta$,where $d$ is the distance between the points and $	heta$ is the angle between the electric field and the displacement vector.Since $A$ and $B$ are at the same potential,the line $AB$ is an equipotential line. The electric field $\vec{E}$ must be perpendicular to the equipotential line $AB$. Thus,$\vec{E}$ is parallel to $BC$.The distance between $B$ and $C$ is $d = BC = 4 \, cm = 0.04 \, m$.Since $\vec{E}$ is parallel to $BC$,the potential difference is $\Delta V = E \cdot d$.$5 \, V = E 	imes 0.04 \, m$.$E = \frac{5}{0.04} \, V/m = 125 \, V/m$ (or $N/C$).

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