Question

A negative charge 6×10^-6c experts attractive force of 65n on a second charge 0.05m away. what is the magnitude and sign of the second charge?

Answer 1

Answer:

The required second charge is +3.01 × 10⁻⁶ C.

Explanation:

Given :

• Charge, q₁ = -6 × 10⁻⁶ C
• Force, F = 65 N
• Distance between the two charges, d = 0.05 m

To find :

the magnitude and sign of the second charge

Solution :

Let the second charge be q₂.

$\sf F = k \dfrac{q1q2}{d^{2}} \\\\ \sf where \ k = 9 \times 10^{9} \\\\ \sf 65 = 9 \times 10^{9} \bigg(\dfrac{6 \times 10^{-6} \times q_2}{0.05^{2}}\bigg) \\\\ \sf q_2 = \dfrac{ 65(25 \times 10^{-4})}{54 \times 10^{3}} \\\\ \sf q_2 = 30.09 \times 10^{-7} \\\\ \sf q_2=3.01 \times 10^{-6} \: C$

Since the force is attractive force, the second charge, q₂ must be positive (unlike charges attract each other)

Answer 2

Answer:

The required second charge is +3.01 × $10^{-6}$ C.

Explanation:

Given :

Charge, q₁ = -6 × $10^{-6}$ C

Force, F = 65 N

Distance between the two charges, d = 0.05 m

To find :

The magnitude and sign of the second charge

Solution :

Let the second charge be q₂.

According to Coulomb's Law of Electrostatics, the force (F) between two charged particles, $q_{1}$ and $q_{2}$, separated by distance d is given by,

$F =k \frac{q_{1} q_{2} }{d^{2} }$

where k is the Coulomb's Constant whose value is equal to 9 × $10^{9}$

Now, When we replace the values in the equation, thus we obtain

$65 =9 \times 10^{9}\frac{(6 \times10^{-6} \times q_{2}) }{(0.05)^{2} }$

$q_{2} = \frac{65(25 \times 10^{-4}) }{54 \times 10^{3} } = 30.09 \times 10^{-7} = 3.01 \times 10^{-6} C$

So, the value of the second charge is 3.01 × $10^{-6}$

Therefore, the magnitude and sign of the second charge is +3.01 × $10^{-6}$ C.

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