Electrostatics 2 Question 4
4. The electric field in a region is given by $\mathbf{E}=(A x+B) \hat{\mathbf{i}}$, where $E$ is in $NC^{-1}$ and $x$ is in metres. 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 (Main 2019, 8 April II)
(a) $-48 V$
(b) $-520 V$
(c) $180 V$
(d) $320 V$
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Answer:
Correct Answer: 4. (c)
Solution:
- Given, $\mathbf{E}=(A x+B) \hat{\mathbf{i}} N-C^{-1}$
The relation between electric field and potential is given as
$$ d V=-\mathbf{E} \cdot d \mathbf{x} $$
Integrating on both sides within the specified limits, we get
$$ \begin{aligned} \therefore \quad \int _1^{2} d V & =V _2-V _1=-\int _{x _1}^{x _2} \mathbf{E} \cdot d \mathbf{x} \\ \Rightarrow \quad V _1-V _2 & =\int _{x _1}^{x _2} \mathbf{E} \cdot d \mathbf{x} \\ & =\int _{x _1}^{x _2}(A x+B) \hat{\mathbf{i}} \cdot(d x \hat{\mathbf{i}})=\int _{x _1}^{x _2}(A x+B) \cdot d x \end{aligned} $$
Here, $A=20$ SI unit, $B=10$ SI unit,
$$ \begin{aligned} x _1=1 \text { and } x _2 & =-5 \\ \Rightarrow \quad V _1-V _2 & =\int _1^{-5}(20 x+10) \cdot d x \\ & =\frac{20 x^{2}}{2}+10 x=10\left[x^{2}+x\right] _1^{-5} \\ & =10\left[(-5)^{2}+(-5)-(1)^{2}-(1)\right] \\ & =10(25-5-2)=180 V \end{aligned} $$
