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2. Electric Potential and Capacitance
medium
The election field in a region is given by $\vec E = (Ax + B)\hat i$ where $E$ is in $N\,C^{-1}$ and $x$ in meters. The values of constants are $A = 20\, SI\, unit$ and $B = 10\, SI\, unit$. If the potential at $x =1$ is $V_1$ and that at $x = -5$ is $V_2$ then $V_1 -V_2$ is.....$V$
A
$320$
B
$-48$
C
$-520$
D
$180$
(JEE MAIN-2019)
Solution
$\overrightarrow{\mathrm{E}}=(20 x+10) \hat{\mathrm{i}}$
$\mathrm{V}_{1}-\mathrm{V}_{2}=-\int_{-5}^{1}(20 \mathrm{x}+10) \mathrm{d} \mathrm{x}$
$\mathrm{V}_{1}-\mathrm{V}_{2}=-\left(10 \mathrm{x}^{2}+10 \mathrm{x}\right)_{-5}^{1}$
$\mathrm{V}_{1}-\mathrm{V}_{2}=10(25-5-1-1)$
$V_{1}-V_{2}=180\, V$
Standard 12
Physics
