Question

Three equal charges of 4 x 10-7 C are located at the corners of a right angled triangle
ABC whose sides are AB – 6 cm, BC = 8 cm and CA = 10 cm. Find the force exerted on
the charge located at the 90° angle. [8 marks

Answer 1

Answer :

• The force exerted (Resultant force) on the charge located at the 90° angle is 0.46 N

Explanation :

Given :

• Magnitude of three equal charges, Q = 4 × 10⁻⁷ C

• Side of the triangle, AB = 6 cm or 6 × 10⁻² m

• Side of the triangle, BC = 8 cm or 8 × 10⁻² m

• Side of the triangle, AC = 10 cm or 10 × 10⁻² m

To find :

• Force exerted on the charge located at 90° angle, Fr = ?

Knowledge required :

Formula for force :

$\boxed{\sf{F = \dfrac{1}{4\pi\epsilon}\dfrac{Qq}{r^{2}}}}$

Where :

• $\bf{\dfrac{1}{4\pi\epsilon}}$ = Permittivity of a medium, here the constant.

• Q/q = Positive point charges.

• r = Distance between the charge

The value of $\bf{\dfrac{1}{4\pi\epsilon}}$ is taken as 9 × 10⁹N m²c⁻²

Solution :

Now,

Let's find out the force exerted by A on B .

Here,

• Q = 4 × 10⁻⁷ C
• q = 4 × 10⁻⁷ C
• r = 6 × 10⁻² m

By using the formula for force and substituting the values in it, we get :

$:\implies \sf{F_{1} = \dfrac{1}{4\pi\epsilon}\dfrac{Qq}{r_{1}^{2}}}$

$:\implies \sf{F_{1} = 9 \times 10^{9} \times \dfrac{4 \times 10^{-7} \times 4 \times 10^{-7}}{(6 \times 10^{-2})^{2}}}$

$:\implies \sf{F_{1} = 9 \times 10^{9} \times \dfrac{16 \times 10^{(-7) + (-7)}}{(6 \times 10^{-2})^{2}}}$

$:\implies \sf{F_{1} = 9 \times 10^{9} \times \dfrac{16 \times 10^{(-14)}}{(6 \times 10^{-2})^{2}}}$

$:\implies \sf{F_{1} = 9 \times 10^{9} \times \dfrac{16 \times 10^{(-14)}}{36 \times 10^{-4}}}$

$:\implies \sf{F_{1} = 9 \times 10^{9} \times \dfrac{16 \times 10^{(-14) - (-4)}}{36}}$

$:\implies \sf{F_{1} = 9 \times 10^{9} \times \dfrac{16 \times 10^{(-14) + 4}}{36}}$

$:\implies \sf{F_{1} = 9 \times 10^{9} \times \dfrac{16 \times 10^{(-10)}}{36}}$

$:\implies \sf{F_{1} = 9 \times \dfrac{16 \times 10^{(-10) + 9}}{36}}$

$:\implies \sf{F_{1} = 9 \times \dfrac{16 \times 10^{(-1)}}{36}}$

$:\implies \sf{F_{1} = \dfrac{16 \times 10^{(-1)}}{4}}$

$:\implies \sf{F_{1} = 4 \times 10^{(-1)}}$

$:\implies \sf{F_{1} = 0.4}$

$\boxed{\therefore \sf{F_{1} = 0.4\:N}}$

Hence the force exerted by A on B is 0.4 N.

Let's find out the force exerted by B on C .

Here,

• Q = 4 × 10⁻⁷ C
• q = 4 × 10⁻⁷ C
• r = 8 × 10⁻² m

By using the formula for force and substituting the values in it, we get :

$:\implies \sf{F_{2} = \dfrac{1}{4\pi\epsilon}\dfrac{Qq}{r_{2}^{2}}}$

$:\implies \sf{F_{2} = 9 \times 10^{9} \times \dfrac{4 \times 10^{-7} \times 4 \times 10^{-7}}{(8 \times 10^{-2})^{2}}}$

$:\implies \sf{F_{2} = 9 \times 10^{9} \times \dfrac{16 \times 10^{(-7) + (-7)}}{(8 \times 10^{-2})^{2}}}$

$:\implies \sf{F_{2} = 9 \times 10^{9} \times \dfrac{16 \times 10^{(-14)}}{(8 \times 10^{-2})^{2}}}$

$:\implies \sf{F_{2} = 9 \times 10^{9} \times \dfrac{16 \times 10^{(-14)}}{64 \times 10^{-4}}}$

$:\implies \sf{F_{2} = 9 \times 10^{9} \times \dfrac{16 \times 10^{(-14) - (-4)}}{36}}$

$:\implies \sf{F_{1} = 9 \times 10^{9} \times \dfrac{16 \times 10^{(-14) + 4}}{64}}$

$:\implies \sf{F_{2} = 9 \times 10^{9} \times \dfrac{16 \times 10^{(-10)}}{64}}$

$:\implies \sf{F_{2} = 9 \times \dfrac{16 \times 10^{(-10) + 9}}{64}}$

$:\implies \sf{F_{2} = 9 \times \dfrac{16 \times 10^{(-1)}}{64}}$

$:\implies \sf{F_{2} = \dfrac{144 \times 10^{(-1)}}{64}}$

$:\implies \sf{F_{2} = 2.25 \times 10^{(-1)}}$

$:\implies \sf{F_{2} = 0.225}$

$\boxed{\therefore \sf{F_{2} = 0.225\:N}}$

Hence the force exerted by B on C is 0.225 N.

Now we know the formula for resultant force i.e,

$\boxed{\bf{F_{r} = \sqrt{F_{1}^{2} + F_{2}^{2}}}}$

Where :

• Fr = Resultant force
• F1 / F2 = Force exerted

Now by using the formula for resultant force and substituting the values in it, we get :

$:\implies \bf{F_{r} = \sqrt{F_{1}^{2} + F_{2}^{2}}}$

$:\implies \bf{F_{r} = \sqrt{0.16 + 0.05}}$

$:\implies \bf{F_{r} = \sqrt{0.21}}$

$:\implies \bf{F_{r} = 0.46(approx.)}$

$\boxed{\therefore \bf{F_{r} = 0.46\:N}}$

Therefore,

• The force exerted (Resultant force) on the charge located at the 90° angle is 0.46 N