Question 101: Electrostats & Capacitance

Question

Two point charges $$20 \mu \mathrm{C}$$ and $$-10 \mu \mathrm{C}$$ are separated by a distance of $$1 \mathrm{~m}$$ in air. At what point on the line joining the two charges, the electric potential is zero.

Accepted Answer

To determine at which point along the line joining the two charges the electric potential is zero, we need to understand how electric potential due to a point charge is calculated and then apply this principle to both charges.
The electric potential, $ V $, at a point in space due to a point charge, $ q $, is given by:
$ V = \frac{k \cdot q}{r} $
where
$ V $ is the electric potential,
$ k $ is Coulomb's constant ($ k = 9 	imes 10^9 \, \mathrm{Nm^2/C^2} $),
$ q $ is the magnitude of the charge, and
$ r $ is the distance from the charge to the point at which the potential is being measured.
Given two charges, $ q_1 = 20\, \mu C = 20 	imes 10^{-6}\, C $ and $ q_2 = -10\, \mu C = -10 	imes 10^{-6}\, C $, separated by $ 1\, m $, to have a net electric potential of zero, the electric potentials due to each charge at a point must be equal in magnitude and opposite in sign.
Let's say the point where the potential becomes zero is at a distance $ x $ from the $ 20\, \mu C $ charge along the line joining the charges. Consequently, this point is $ 1 - x $ meters from the $ -10\, \mu C $ charge.
Setting the absolute values of the potentials equal to each other gives us:
$ \frac{k \cdot |20 	imes 10^{-6}|}{x} = \frac{k \cdot |-10 	imes 10^{-6}|}{1 - x} $
Since $ k $ and the charge magnitudes $ |20 	imes 10^{-6}| $ and $ |-10 	imes 10^{-6}| $ are constant, we can simplify this to:
$ \frac{20}{x} = \frac{10}{1 - x} $
Solving for $ x $:
$ 20(1 - x) = 10x $
$ 20 - 20x = 10x $
$ 20 = 30x $
$ x = \frac{20}{30} = \frac{2}{3} $
Therefore, $ x = 0.67 \, m $.
This means that the electric potential is zero at a point $ 0.67\, m $ from the $ 20\, \mu C $ charge.
The correct answer is:
Option C: $ 0.67\, m $ from the charge $ 20\, \mu C $.

Answer

$$0.5 \mathrm{~m}$$ from the charge $$10 \mu \mathrm{C}$$

Answer

$$0.76 \mathrm{~m}$$ from the charge $$20 \mu \mathrm{C}$$

Answer

$$0.67 \mathrm{~m}$$ from the charge $$20 \mu \mathrm{C}$$

Answer

$$0.25 \mathrm{~m}$$ from the charge $$10 \mu \mathrm{C}$$

Electrostats & Capacitance

Two point charges $20 \mu \mathrm{C} and -10 \mu \mathrm{C} are separated by a distance of 1 \mathrm{~m}$ in air. At what point on the line joining the two charges, the electric potential is zero.

Choose the correct answer: