AP EAMCET Engineering 23 Apr 2019 Shift 1 Paper

The electrostatic force between charges is,
F1=

kQ1 Q2

R2

F1=

k

R2

(3.72×1.86)10−12N
F1=

k

R2

(6.9192×10−12)
If 20% of Q1 is given to Q2 then,
Q′2=Q2+

20

100

Q1
Q′2=1.86+

20

100

(3.72)
Q′2=2.604µC
And
Q′1=

80

100

Q1
Q′1=2.976µC
The new electrostatic force between charges is,
F2=

kQ′1Q′2

R2

F2=

k

R2

(2.976×2.604)10−12N
F2=

k

R2

(7.7495)
The % increases in electrostatic force between charges is,

F2−F1

F1

×100 =

k

R2

(7.7495−6.9192)×10−12

k R2 ×6.9192×10−12

×100
=12%