A point charge of +5 µC is placed at the origin. What is the electric potential at a point 2 m away from the charge?
Practice Questions
1 question
Q1
A point charge of +5 µC is placed at the origin. What is the electric potential at a point 2 m away from the charge?
1125 V
450 V
225 V
0 V
The electric potential V at a distance r from a point charge Q is given by V = kQ/r. Here, V = (9 x 10^9 Nm²/C²)(5 x 10^-6 C)/(2 m) = 1125 V.
Questions & Step-by-step Solutions
1 item
Q
Q: A point charge of +5 µC is placed at the origin. What is the electric potential at a point 2 m away from the charge?
Solution: The electric potential V at a distance r from a point charge Q is given by V = kQ/r. Here, V = (9 x 10^9 Nm²/C²)(5 x 10^-6 C)/(2 m) = 1125 V.
Steps: 6
Step 1: Identify the values given in the problem. We have a point charge Q = +5 µC (which is 5 x 10^-6 C) and the distance r = 2 m.
Step 2: Recall the formula for electric potential V due to a point charge, which is V = kQ/r, where k is the electrostatic constant (approximately 9 x 10^9 Nm²/C²).
Step 3: Substitute the values into the formula. We have V = (9 x 10^9 Nm²/C²) * (5 x 10^-6 C) / (2 m).
Step 4: Calculate the numerator: (9 x 10^9) * (5 x 10^-6) = 45,000.
Step 5: Now divide the result by the distance: 45,000 / 2 = 22,500.
Step 6: Therefore, the electric potential V at a point 2 m away from the charge is 22,500 V.
