Physics, asked by makaljarh2002, 1 month ago

A parallel plate capacitor, with a capacitance
of 2 x 10-6 F, is connected across a 19.1 V
battery. If, with the battery still connected,
you pull the plates apart until their
separation is now twice of what it originally
was, then the magnitude of the charge on
each plate, in µC, is:​

Answers

Answered by prathamesh299
1

Answer:

PLS MARK BRAINLIEST & ♥️⭐

Explanation:

C=dϵoA

According to the values given C=1.8×10−11F

Now Q=CV

Hence charge on the positive plate will be 1.8×10−9C

(a) Dielectric constant, k = 6

C=1.8×10−11 F

new capacitance, C′=kC=108 pF

V=100volts

q′=V×C′=1.08×10−8 C

V remains the same.

(b) C′=kC=6×1.8×10−11=108pF

q remains the same.

q=1.8×10−9 C

V′=q/C′=16.67 V

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