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If the electric potential at a point is 50 V and the electric field strength is

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Question: 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?

Options:

  1. 10 m
  2. 5 m
  3. 25 m
  4. 50 m

Correct Answer: 10 m

Solution: 

Distance d = V/E = 50 V / 5 N/C = 10 m.

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?
  1. 10 m
  2. 5 m
  3. 25 m
  4. 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.
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