If the electric potential at a point is 50 V and the electric field strength is
Practice Questions
Q1
If the electric potential at a point is 50 V and the electric field strength is 5 N/C, what is the distance from the point to the reference point where the potential is zero?
10 m
5 m
25 m
50 m
Questions & Step-by-Step Solutions
If the electric potential at a point is 50 V and the electric field strength is 5 N/C, what is the distance from the point to the reference point where the potential is zero?
Step 1: Understand the problem. We have an electric potential (V) of 50 volts and an electric field strength (E) of 5 newtons per coulomb (N/C).
Step 2: Recall the formula to find the distance (d) from the point to the reference point where the potential is zero. The formula is d = V / E.
Step 3: Substitute the values into the formula. Here, V = 50 V and E = 5 N/C.
Step 4: Calculate the distance. Divide 50 V by 5 N/C: d = 50 / 5.
Step 5: Perform the division. 50 divided by 5 equals 10.
Step 6: Conclude that the distance from the point to the reference point where the potential is zero is 10 meters.
Electric Potential and Electric Field – The relationship between electric potential (V), electric field strength (E), and distance (d) is given by the formula d = V/E, where V is the potential difference and E is the electric field strength.
