Charges of 12 nC and 16 nC at the ends of 6 cm legs, a 1 nC charge at the right angle. Find the force on it — Physics problems | PhysLabo
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Force on a charge at the right angle from two charges

Medium 10 grade Electricity / Electrostatics
Problem statement

Positive point charges $q_1 = 12$ nC and $q_2 = 16$ nC are placed at vertices $A$ and $B$ of a right triangle $ABC$ with the right angle at $C$ and equal legs $CA = CB = a = 6$ cm. A positive point charge $q_0 = 1$ nC is placed at vertex $C$ (see the figure). Find the magnitude of the force acting on the charge $q_0$.

A right triangle: charges q₁ and q₂ at the acute vertices, charge q₀ at the right angle, legs equal to a
Given
  • $q_1 = 12$ nC $= 12 \cdot 10^{-9}$ C
  • $q_2 = 16$ nC $= 16 \cdot 10^{-9}$ C
  • $q_0 = 1$ nC $= 10^{-9}$ C
  • $a = 6$ cm $= 0.06$ m
  • $k = 9 \cdot 10^9$ N·m²/C²
  • $F = ?$
Hints
Hint 1 / 2
Like charges repel: $\vec F_1$ points away from $q_1$ and $\vec F_2$ away from $q_2$. These vectors are perpendicular to each other.
Hint 2 / 2
Find $F_1 = kq_0q_1/a^2$ and $F_2 = kq_0q_2/a^2$ separately, then add the vectors with the Pythagorean theorem.
Solution

By the superposition principle the force on $q_0$ is the vector sum of the forces $\vec F_1$ and $\vec F_2$ from each charge. All the charges are positive, so the forces are repulsive: $\vec F_1$ points along $AC$ away from $q_1$, $\vec F_2$ along $BC$ away from $q_2$. These forces are perpendicular.

Adding the forces by the parallelogram rule
Adding the forces by the parallelogram rule

1. Magnitudes from Coulomb's law:

$$F_1 = k\frac{q_0 q_1}{a^2} = 9 \cdot 10^9 \cdot \frac{10^{-9} \cdot 12 \cdot 10^{-9}}{0.06^2} = 3 \cdot 10^{-5} \text{ N}$$$$F_2 = k\frac{q_0 q_2}{a^2} = 9 \cdot 10^9 \cdot \frac{10^{-9} \cdot 16 \cdot 10^{-9}}{0.06^2} = 4 \cdot 10^{-5} \text{ N}$$

2. Add the perpendicular vectors with the Pythagorean theorem:

$$F = \sqrt{F_1^2 + F_2^2} = \sqrt{(3\cdot10^{-5})^2 + (4\cdot10^{-5})^2} = 5 \cdot 10^{-5} \text{ N} = 50\ \mu\text{N}$$

Direction: $\tan\varphi = F_2/F_1 = 4/3$, so $\varphi \approx 53°$ to side $CA$ (pointing away from the triangle).

Answer

$F = 5 \cdot 10^{-5}$ N $= 50\ \mu$N

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