Two charges 14 μC and −4 μC are placed at (−12 cm,0,0) and (12 cm,0,0) in an external electric field E=(Br2), where B=1.2×106 N/cm2 and r is in metres. The electrostatic potential energy of the configuration is
Question
Two charges 14 μC and −4 μC are placed at (−12 cm,0,0) and (12 cm,0,0) in an external electric field E=(Br2), where B=1.2×106 N/cm2 and r is in metres. The electrostatic potential energy of the configuration is
Solution
The electrostatic potential energy of a system of charges is given by the formula:
U = k * q1 * q2 / r
where:
- U is the potential energy
- k is Coulomb's constant (8.99 * 10^9 N m^2/C^2)
- q1 and q2 are the charges
- r is the distance between the charges
In this case, q1 = 14 μC = 14 * 10^-6 C, q2 = -4 μC = -4 * 10^-6 C, and r = 24 cm = 0.24 m (since the charges are 12 cm apart from the origin in opposite directions).
So, the potential energy of the system is:
U = 8.99 * 10^9 N m^2/C^2 * 14 * 10^-6 C * -4 * 10^-6 C / 0.24 m U = -209.58 J
Now, the potential energy due to the external electric field E = Br^2 is given by:
U = q * E * r
where E = Br^2 = 1.2 * 10^6 N/C m^2 * (0.24 m)^2 = 6.912 * 10^6 N/C.
So, the potential energy of the system due to the external field is:
U = 14 * 10^-6 C * 6.912 * 10^6 N/C * 0.24 m - 4 * 10^-6 C * 6.912 * 10^6 N/C * 0.24 m U = 23.17 J - 6.65 J U = 16.52 J
Therefore, the total potential energy of the configuration is the sum of the potential energy due to the charges and the potential energy due to the external field:
U_total = -209.58 J + 16.52 J U_total = -193.06 J
So, the electrostatic potential energy of the configuration is -193.06 J.
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