Two positive charges of 0.2microC and 0.01 microC are placed 10cm apart. Calculate the work done in reducing their distance to 5 cm
2 answers
1.8x10*-2
The work done in reducing the distance between the charges from 10 cm to 5 cm is:
W = K * q1 * q2 * (1/r2 - 1/r1)
where:
K = Coulomb's constant = 9x10^9 N*m^2/C^2
q1 = 0.2 microC = 2x10^-7 C
q2 = 0.01 microC = 1x10^-8 C
r1 = initial distance = 10 cm = 0.1 m
r2 = final distance = 5 cm = 0.05 m
Plugging in the values, we get:
W = 9x10^9 * 2x10^-7 * 1x10^-8 * (1/0.052 - 1/0.12) = 1.8x10^-2 J
Therefore, the work done in reducing their distance to 5 cm is 1.8x10^-2 J.
W = K * q1 * q2 * (1/r2 - 1/r1)
where:
K = Coulomb's constant = 9x10^9 N*m^2/C^2
q1 = 0.2 microC = 2x10^-7 C
q2 = 0.01 microC = 1x10^-8 C
r1 = initial distance = 10 cm = 0.1 m
r2 = final distance = 5 cm = 0.05 m
Plugging in the values, we get:
W = 9x10^9 * 2x10^-7 * 1x10^-8 * (1/0.052 - 1/0.12) = 1.8x10^-2 J
Therefore, the work done in reducing their distance to 5 cm is 1.8x10^-2 J.