Question 1 of 11intermediate🔧 ApplyNumerical2 marks

Two charges 5×105 \times 10–8 C and –3×103 \times 10–8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Correct Answer

The electric potential is zero at two points: (i) 10 cm from the 5×105 \times 10⁻⁸ C charge (between the two charges), and (ii) 40 cm from the 5×105 \times 10⁻⁸ C charge (on the side of the negative charge, away from both charges).

Exercise: EXERCISES | Q: 2.1 | (Chapter: 79)
For More Understanding

Explanation

This question is similar to Example 2.2 in the textbook. Since we have a positive and a negative charge, the potential can be zero at two locations: between the charges (where potentials have opposite signs and can cancel) and outside the charges on the side of the smaller magnitude charge. The potential at infinity is taken as zero, as stated in the problem and consistent with Equation 2.8 in the context.

Solution Steps

  1. Step 1: Let the positive charge q₁ = 5×105 \times 10⁻⁸ C be at origin O. The negative charge q₂ = -3×103 \times 10⁻⁸ C is at 16 cm on the x-axis.

  2. Step 2: For potential to be zero: V₁ + V₂ = 0, which gives kq₁/r₁ + kq₂/r₂ = 0, or q₁/r₁ = -q₂/r₂ = |q₂|/r₂

  3. Step 3: Case I (between charges): Let point P be at distance x from q₁. Then r₁ = x and r₂ = (16 - x). So 5/x = 3/(16-x). Solving: 5(16-x) = 3x, giving x = 10 cm.

  4. Step 4: Case II (outside, on negative charge side): Let P be at distance x from q₁ (x > 16). Then r₁ = x and r₂ = (x - 16). So 5/x = 3/(x-16). Solving: 5(x-16) = 3x, giving x = 40 cm.

  5. Step 5: No solution exists on the positive charge side (x < 0) as both potentials would be positive there.

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