Two small identical metal balls charged to 5 micro C and -15 micro C are placed 0.1 m apart in air. Find the force between them . What will be the force if they are allowed to touch each other and separated as before ?
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q1= 5 x 10^-6 C
q2=-15 x 10^-6 C
F= -9 x 10^9 x 45 x 10^-12 / 0.01
F= -405 x 10^-3 / 0.01
F= -40500 x 10^-3
F=- 40.5 N/ C
Now ,if they are touched and separate again .
The value of new charge on both body is same .
= ( -15 + 5) /2
= -5
F' = K x 25 x 10^-12 /0.01
F' = 9 x10^9 x 25 x 10^-12 /0.01
F'= 9 x 25 x10^-3 /0.01
F'= 225 x 10^-1
F'= 22.5 N/ C
Hope it helps you : ]
____________________________________________________________________________________________________________________________
q1= 5 x 10^-6 C
q2=-15 x 10^-6 C
F= -9 x 10^9 x 45 x 10^-12 / 0.01
F= -405 x 10^-3 / 0.01
F= -40500 x 10^-3
F=- 40.5 N/ C
Now ,if they are touched and separate again .
The value of new charge on both body is same .
= ( -15 + 5) /2
= -5
F' = K x 25 x 10^-12 /0.01
F' = 9 x10^9 x 25 x 10^-12 /0.01
F'= 9 x 25 x10^-3 /0.01
F'= 225 x 10^-1
F'= 22.5 N/ C
Hope it helps you : ]
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