Two points charged Q1 =2uc and Q2= _4uc are placed on the x-axis at x= 60cm and x =30cm respectively what is the potential energy of the tests charged q =6uc placed at x= 60cm

Answer 1

To calculate the potential energy of the test charge q = 6 uC placed at x = 60 cm, we can use the formula for potential energy of a system of charges:

U = k * (q1 * q) / r1 + k * (q2 * q) / r2

Where:
U = potential energy
k = Coulomb's constant = 8.9875 x 10^9 N m^2 C^-2
q1 = charge of Q1 = 2 uC = 2 x 10^-6 C
q2 = charge of Q2 = -4 uC = -4 x 10^-6 C
q = test charge = 6 uC = 6 x 10^-6 C
r1 = distance between Q1 and q = 60 cm = 0.6 m
r2 = distance between Q2 and q = 30 cm = 0.3 m

Now, we can substitute these values into the formula:

U = (8.9875 x 10^9) * {(2 x 10^-6 * 6 x 10^-6) / 0.6} + (8.9875 x 10^9) * {(-4 x 10^-6 * 6 x 10^-6) / 0.3}

Calculating the value of this expression gives:

U = (8.9875 x 10^9) * {12 x 10^-12 / 0.6} + (8.9875 x 10^9) * {-24 x 10^-12 / 0.3}
U = (8.9875 x 10^9) * 2 x 10^-12 + (8.9875 x 10^9) * -8 x 10^-12
U = 1.7975 x 10^-2 - 7.18 x 10^-2
U = 0.7185 x 10^-2
U = 7.185 x 10^-3 J

Therefore, the potential energy of the test charge q = 6 uC placed at x = 60 cm is 7.185 x 10^-3 Joules.