Question

A charge +q exerts a force of magnitude -0.2N on another charge -2q. If they are separated by 36 cm. Determine the value of q

1) 2.2 micro-coulomb
2) 1.2 micro-coulomb
3)1.18 micro-coulomb
4) 12 micro-coulomb

Answer 1

Given :

• 1st Charge = +q
• 2nd charge = -2q
• Distance between them = 36cm
• Electrostatic Force between them = -0.2N

To find :

Value of q

Formula used :

$\sf \: F_e \: = \dfrac{k \: \times q_{1} \times \: q_{2} }{ {r}^{2} }$

Here,

• Fₑ = Electrostatic Force
• q₁ and q₂ are charge on two bodies
• r is distance between them

Solution :

Putting value in above formula we get,

$\sf \: \implies \: - 0.2N \: = \dfrac{9 \times 10^9 \: \times q \times \: (-2q) }{ {(36cm)}^{2} }$

$\sf \: \implies \: - 0.2N \: = \dfrac{9 \times 10^9 \: \times \: - 2 {q}^{2} }{ {(0.36m)}^{2} }$

$\sf \: \implies \: {q}^{2} \: = \: \dfrac{ - 0.2N \: \times {(0.36m)}^{2} }{9 \times 10^9 \: \times \: - 2 }$

$\sf \: \implies \: {q}^{2} \: = \: \dfrac{ 0.1 \: \times 0.1296 }{9 \times 10^9 \: } {C}^{2}$

$\sf \: \implies \: {q}^{2} \: = \: \dfrac{ 0.01296 }{9 \times 10^9 \: } {C}^{2}$

$\sf \: \implies \: {q}^{2} \: = \: \dfrac{ 0.00144}{ 10^9 \: } {C}^{2}$

$\sf \: \implies \: {q}^{2} \: = \: \dfrac{ 144 \times {10 }^{ - 5} }{ 10^9 \: } \: {C}^{2}$

$\sf \: \implies \: {q}^{2} \: = \: 144 \times {10 }^{( - 5 - 9)} {C}^{2}$

$\sf \: \implies \: {q}^{2} \: = \: 144 \times {10 }^{( - 14)} {C}^{2}$

$\sf \: \implies \: {q}\: = \sqrt{144 \times {10 }^{( - 14)} } \: C$

$\sf \: \implies \: {q}\: = \pm \: 12 \times {10}^{ - 7} C$

$\sf \: \implies \: {q}\: = \pm \: 1.2 \times {10}^{ - 6} C$

$\sf \: \implies \: \bold{{q}\: = \pm \: 1.2 \: micro C}$

Note - As in option we are given only positive values, so

$\sf \: \implies \: \: \bold{{q}\: = \: 1.2 \: micro C}$

ANSWER :

option 2) 1.2 micro Coulomb