Question

Find the distance at which the total potential will be zero for an arrangement of two charges each of magnitude of 4 μC and -1μC and is separated by a distance of 10 cm in a straight line.
1. 8 cm
2. 7 cm
3. 6 cm
4. 5 cm

Answer 1

Correct Answer - Option 1 : 8 cm

CONCEPT: 

• The amount of work done in moving a unit positive charge in an electric field from infinity to that point without accelerating the charge against the direction of the electric field. The electrostatic potential is given by

\(⇒ V = \frac{W}{q}\)

• The electrostatic potential due to a point charge is given by 

\(⇒ V = \frac{Kq}{r}\)                                        \([ \because K = \frac{1}{4\pi\epsilon_{0}}]\)

Where W = Work , q = charge, r = distance

• The expression for total potential due two charges q₁ and q₂ is given by

\(⇒ V_{total} = K(\frac{q_{1}}{r_{1}} +\frac{q_{2}}{r_{2}})\)

Where q₁ = q₂ = Charges, and r₁ = r₂ = Distance

CALCULATION:

Given - q₁ = 4 μc, q₂ = -1μc, r = 10 cm, r₁ = X

Let x be the distance away from the charge 4μc where the potential experienced is zero (r1).

⇒ r2 = 10 - X

• The total potential of the system is given by 

\(⇒ V_{total} = K(\frac{q_{1}}{r_{1}} +\frac{q_{2}}{r_{2}})\)

Substituting the given values and conditions in the above equation

\(⇒ 0 = K(\frac{4\mu c}{X} -\frac{1 \mu c}{(10-X)})\)

\(⇒ 0 = (\frac{4\mu c}{X} -\frac{1 \mu c}{(10-X)})\)

\(⇒ \frac{4\mu c}{X} = \frac{1\mu c}{(10 -X)}\)

\(⇒ 5X = 40 \)

\(⇒ X = 8 cm\)

• Hence option 1 is the answer