Question

In uniform electric field $E=10\text{N/C}$ as shown in figure. The potential difference, $V_A-V_B$ and $V_B-V_C$ respectively

• A. $10\text{V},-20\text{V}$
• B. $-10\text{V},20\text{V}$
• C. $-10\text{V},-20\text{V}$
• D. $10\text{V},20\text{V}$

Answer 1

The correct option is B $-10\text{V},20\text{V}$

Given,
Intensity of electric field, $E=10\text{NC}$

Since, electric field is directed from higher electric potential to lower electric potential means electric potential decreases in the direction of electric field.

Thus, $V_B>V_A$, So, $(V_A-V_B)$ will be negative.

From figure, the distance $d_{AB}$ between the points $\text{A}$ and $\text{B}$ along $E$,

$d_{AB}=2\cos {60}^{\circ }=1\text{m}$

Using the relation, $V_A-V_B=-E{d}_{AB}$
we get,

$V_A-V_B=(-10)\left(1\right)$

$\Rightarrow V_A-V_B=-10\text{V}$

Similarly, for uniform $\overrightarrow{E}$

$V_B>V_C$, So $(V_B-V_C)$ will be positive.

From figure, the distance $d_{BC}$ between the points $\text{B}$ and $\text{C}$ along $E$,

$d_{BC}=2.0\text{m}$

Using the relation, $V_B-V_C=E{d}_{BC}$
we get,

$V_B-V_C=\left(10\right)\left(2\right)$

$\therefore V_B-V_C=20\text{V}$

Hence, option (b) is correct answer.