two charges 4cand. 16 c separated by a distance 30cm. what is the value of third charge 4c? at which net force is equal to zero where the vlue of third charge is 20c
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Answer:
We know that,
According to coulomb's law,the magnitude of the force of attraction or repulsion between two point charges is directly proportional to the product of magnitude of the charges and inversely proportional to the square of the distance between them.. Therefore
F = k Q1Q2/r^2
Therefore, Q1Q2 = Fr^2/ k
Q1Q2 = 0.075 * 3^2/8.99*10^9
Q1Q2 = 7.5 * 10^-11
Now,
According to d condition,
Q1 + Q2 = 20 * 10^ (-6)
Put it in equation 1 after solving we get d value of Q1 which is equal to 5 * 10^ (-6) C
And similarly,Q2 will be 15 *10^ (-6)
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