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:
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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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