Question 56: Electrostats & Capacitance

Question

A dielectric of dielectric constant $$K$$ is introduced such that half of its area of a capacitor of capacitance $$C$$ is occupied by it. The new capacity is

Accepted Answer

The dielectric is introduced such that half of its area is occupied by it.
In the given case, the two capacitors are in parallel.
$$	herefore \quad C^{\prime}=C_1+C_2$$
But $$C_1=\frac{A \varepsilon_0}{2 d} 	ext { and } C_2=\frac{K A \varepsilon_0}{2 d}$$
Thus, $$C^{\prime}=\frac{A \varepsilon_0}{2 d}+\frac{K A \varepsilon_0}{2 d} \Rightarrow C^{\prime}=\frac{C}{2}(1+K)$$

Answer

$$2 C$$

Answer

$$\frac{C}{2}$$

Answer

$$\frac{(1+K) C}{2}$$

Answer

$$2 C(1+K)$$

Electrostats & Capacitance

A dielectric of dielectric constant $K is introduced such that half of its area of a capacitor of capacitance C$ is occupied by it. The new capacity is

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