AP EAPCET 06-Jul-2022 Shift 2 Paper

Given,
C1‌=C2=8µF=8×10−6F
V‌=10V
Since, C1 and C2 are connected in parallel. Equivalent capacitance,
Total charge, qi=Ceq⋅V
‌=1.6×10−5×10
‌=1.6×10−4C
We know that, capacitance C=‌

ε0A

d

⇒C∝‌

1

d

⇒‌

C′

C′′

=‌

d′′

d′

=‌

(d′−40%‌ of ‌d)

d′

=‌

3∕5d′

d′

⇒‌

Cr

C′′

=‌

3

5

⇒C′′=‌

5

3

C′
‌=‌

5

3

×8×10−6=‌

40

3

×10−6C
New capacitance, C′′′=C′′+C2
‌=‌

40

3

×10−6+8×10−6
‌=‌

64

3

×10−6F
Now value of total charge, qf=C′′′×V
‌=‌

64

3

×10−6×10
‌=‌

64

3

×10−5C
Increased value of charge,
∆q‌=qf−q1
‌=‌

64

3

×10−5−1.6×10−4
‌=‌

160

3

µC