Question

Two charged spheres exert a force of 5.54 × 10-4 N on each other. If one of the spheres has a charge of 3.72 × 10-9 C and the other sphere has a charge of 4.32 × 10-6 C, how far apart are the charged spheres?

Answer 1

Given :

• Electrostatic force between two charged species = 5.54 × 10⁻⁴ N
• Charge on one sphere = 3.72 × 10⁻⁹ C
• Charge on other sphere = 4.32 × 10⁻⁶ C

To find :

Distance between them

Formula used :

Fₑ = K × q₁ × q₂ ÷ ( r² )

Here

• Fₑ = Electrostatic force
• K = Coulomb's constant
• q₁ = Charge on one sphere
• q₂ = Charge on other sphere
• r = distance between them

Solution :

$\sf \implies \: F_e = \dfrac{K \times q_1 \times \: q_2 }{r^2}$

$\sf \implies \: 5.54 \times {10}^{ - 4} = \dfrac{9 \times {10}^{9} \times 3.72 \times {10}^{ - 9} \times \: 4.32 \times {10}^{ - 6} }{r^2}$

$\sf \implies \: {r}^{2} = \dfrac{9 \times {10}^{9} \times 3.72 \times {10}^{ - 9} \times \: 4.32 \times {10}^{ - 6} }{5.54 \times {10}^{ - 4} \: }$

$\sf \implies \: {r}^{2} = \dfrac{144.6336 \times {10}^{(9 - 9 - ( - 4) - 6)} }{5.54 \: }$

$\sf \implies \: {r}^{2} = \dfrac{144.6336 \times {10}^{ - 2} }{5.54 \: }$

$\sf \implies \: {r}^{2} =26.107 \times {10}^{ - 2}$

$\sf \implies \: {r}^{2} \: = \: 0.26107$

$\sf \implies \: {r} \: = \: \: \sqrt{0.26107}$

$\sf \implies \: {r} \: = \: \pm \: 0.51 \: m$

As ,r denotes distance it can't be negative

$\sf \implies \: {r} \: = \: \bold{0.51m \: = \: 51cm}$

ANSWER : 0.51m = 51cm