Question

what will happen to the following when distance between the plates double than the initial but area of cross section remains same (a) capacitance, (b)charge, (c)potential,(d)electric field,(e) total energy stored .Given battery remains connected with the capacitor?​

Answer 1

Answer:

a) Capacitance becomes half

b) Charge becomes half

c) Potential remains the same

d) Electric field becomes double

e) Total energy will become half

Explanation:

As we know that capacitance of parallel plate capacitor is given as

$C = \frac{\epsilon_0 A}{d}$

here we know that

A = area of plates

d = distance between the plates

now here if we change the distance to double of initial value while the area remains the same

So we can see that the capacitance is inversely dependent on the distance between the plates

so we have

i) Capacitance becomes half that of initial value

ii) As we know that

Q = CV

so here charge will also becomes half

iii) Potential will remain the same as battery is connected

iv) as we know that

V = E. d

since the distance is half and potential is constant so electric field becomes double

v) As we know that total energy stored is given as

$U = \frac{1}{2}CV^2$

so here energy stored becomes half of initial value

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Topic : Capacitance

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