Question

charges 5×10-7 c and -2.5×10-7c are placed at the corners of an equilateral triangle of side 5 cm find the electric force on a charge 1.0×10-7c placed on third corner

Answer 1

is the answer correct

Answer 2

Answer:

The electric force on charge $q_{C}$ is 0.12N.

Explanation:

Given:

$q_{A}=5\times 10^{-7} C$

$q_{B}=-2.5\times 10^{-7} C$

$q_{C}=1\times 10^{-7} C$

$r_{CA}=0.05m$

$r_{CB}=0.05m$

θ = 120°

Here,

Let the vertices of the triangle be denoted by A, B and C.

Let the charge be placed at the vertex A is denoted by $q_{A}$.

Let the charge be placed at the vertex B is denoted by $q_{B}$.

Let the charge be placed at the vertex C is denoted by $q_{C}$.

The distance between the charge $q_{A}$ and $q_{C}$ is denoted by $r_{CA}$.

The distance between the charge $q_{B}$ and $q_{C}$ is denoted by $r_{CB}$.

The force on $q_{C}$ due to $q_{A}$ is denoted by $F_{CA}$.

The force on $q_{C}$ due to $q_{B}$ is denoted by $F_{CB}$.

The resultant force of $F_{CA}$ due to $F_{CB}$ is denoted by $F_{R}$.

Now,

By the formula,

$F_{CA} =\frac{Kq_{C}q_{A} }{r_{CA}^{2} }$

$F_{CA} =\frac{9\times 10^{9} \times1\times 10^{-7} \times 5 \times 10^{-7} }{0.05^{2} }$

$F_{CA} =0.18N$

Then,

By the formula,

$F_{CB} =\frac{Kq_{C}q_{B} }{r_{CB}^{2} }$

$F_{CA} =\frac{9\times 10^{9} \times1\times 10^{-7} \times (-2.5) \times 10^{-7} }{0.05^{2} }$

$F_{CB} =0.09N$

Then,

By the equation,

$F_{R} =\sqrt{F_{CA}^{2} +F_{CB}^{2} +2F_{CA} F_{CB}cos\theta }$

$F_{R} =\sqrt{0.18^{2} +0.09^{2} +2\times 0.18\times 0.09cos120 }$

$F_{R} =0.12N$

So, the electric force on charge $q_{C}$ is 0.12N.