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A -0.00325 C charge q1 is placed 5.62 m from a second charge q2. The first charge is repelled with a 48900 N force. What is the value of q2? C

Include the sign of the charge, - or +

User RxRead
by
4.4k points

2 Answers

5 votes

Answer:

-0.05286

Step-by-step explanation:

acellus

User Josh Butler
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4.2k points
2 votes

Answer: q2 = -0.05286

Step-by-step explanation:

Given that

Charge q1 = - 0.00325C

Electric force F = 48900N

The electric field strength experienced by the charge will be force per unit charge. That is

E = F/q

Substitute F and q into the formula

E = 48900/0.00325

E = 15046153.85 N/C

The value of the repelled second charge will be achieved by using the formula

E = kq/d^2

Where the value of constant

k = 8.99×10^9Nm^2/C^2

d = 5.62m

Substitutes E, d and k into the formula

15046153.85 = 8.99×10^9q/5.62^2

15046153.85 = 284634186.5q

Make q the subject of formula

q2 = 15046153.85/ 28463416.5

q2 = 0.05286

Since they repelled each other, q2 will be negative. Therefore,

q2 = -0.05286

User Yevhenii Nadtochii
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4.1k points