Electrostatics 5 Question 38
40. A dielectric slab of thickness $d$ is inserted in a parallel plate capacitor whose negative plate is at $x=0$ and positive plate is at $x=3 d$. The slab is equidistant from the plates. The capacitor is given some charge. As $x$ goes from 0 to $3 d$
(1998, 2M)
(a) the magnitude of the electric field remains the same.
(b) the direction of the electric field remains the same.
(c) the electric potential increases continuously.
(d) the electric potential increases at first, then decreases and again increases.
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Solution:
- The magnitude and direction of electric field at different points are shown in figure. The direction of the electric field remains the same. Hence, option (b) is correct. Similarly, electric lines always flow from higher to lower potential, therefore, electric potential increases continuously as we move from $x=0$ to $x=3 d$.
Therefore, option (c) is also correct. The variation of electric field $(E)$ and potential $(V)$ with $x$ will be as follows
$$ \begin{array}{ll} \text { Because } & E _{O-d}=E _{2 d-3 d} \\ \text { and } & E _{O-d}>E _{d-2 d} \end{array} $$
$$ O A | B C \text { and }(\text { Slope }) _{O A}>(\text { Slope }) _{A B} $$
