The drawing shows a square, each side of which has a length of L = 0.250 m. Two different positive charges q1 and q2 are fixed at the corners of the square. Find the electric potential energy of a third charge q3 = -7.00 x 10-9 C placed at corner A and then at corner B.
figure the Va, then Vb from the original two charges.
PE at corner A= Va*q3
PE at corner B=Vb*q3
To find the electric potential energy of the third charge q3 at corner A and then at corner B, we need to consider the interaction between the charges involved. The electric potential energy (U) of a system of charges is given by the equation:
U = k * (q1 * q3 / r13) + k * (q2 * q3 / r23)
Where:
- U is the electric potential energy
- k is the electrostatic constant (k = 8.99 x 10^9 N m^2/C^2)
- q1 and q2 are the charges at the corners of the square
- q3 is the charge at corner A or B
- r13 and r23 are the distances between q1 and q3, and q2 and q3 respectively.
To calculate the electric potential energy at corner A, we need to substitute the values into the equation:
U(A) = (8.99 x 10^9 N m^2/C^2) * (q1 * q3 / r13) + (8.99 x 10^9 N m^2/C^2) * (q2 * q3 / r23)
Similarly, to calculate the electric potential energy at corner B, we substitute the values into the equation:
U(B) = (8.99 x 10^9 N m^2/C^2) * (q1 * q3 / r13) + (8.99 x 10^9 N m^2/C^2) * (q2 * q3 / r23)
Note that the distances r13 and r23 can be calculated as the diagonal of the square, which can be found using the Pythagorean theorem.
Once you have the values for q1, q2, q3, r13, and r23, you can substitute them into the equations above to calculate the electric potential energy at corner A and corner B.