Question

5. If a dielectric plate of thickness t is placed between the plates of a parallel plate
capacitor of plate distance d, the capacitance becomes half of the original value.
The dielectric constant of the plate will be
2t
2t
t
t
(a)
(b)
(c)
(d)
2d-t
2d+t
d-t
dti
yed
ide V. is​

Answer 1

Answer:

K = $\frac{t}{d+t}$ is the required dielectric constant of the plate .

Explanation:

Given , thickness of a plate = t

distance between the parallel plate capacitor = d

and capacitance become half of the original .

Step1:

Therefore we have $C^{1}$ = $\frac{C}{2}$

where , $C^{1}$ be the initial capacitance and C be the original capacitance

But we have  $C^{1} = \frac{C}{1-\frac{t}{d}(1-\frac{1}{K} ) }$

⇒ $\frac{C}{2} = \frac{C}{1-\frac{t}{d}(1-\frac{1}{K} ) }$

⇒${1-\frac{t}{d}(1-\frac{1}{K} ) } = 2$

⇒$\frac{t}{d} (1-\frac{1}{K} ) }$ = -1

⇒K = $\frac{t}{d+t}$

Final answer :

Hence , the dielectric constant plate will be k =  $\frac{t}{d+t}$

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