Two small charged objects repel each other with a force F whenseparated by a distance d. if the charge on each object is reducedto one-fourth of its original value and the distance between themis reduced to d/2, the force becomes: a) \frac{F}{16} b) \frac{F}{8} c) \frac{F}{4} d) \frac{F}{2} e)F

Maiclubk 2021-02-21 Answered
Two small charged objects repel each other with a force F whenseparated by a distance d. if the charge on each object is reducedto one-fourth of its original value and the distance between themis reduced to d/2, the force becomes:
a) F16
b) F8
c) F4
d) F2
e)F
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Expert Answer

2k1enyvp
Answered 2021-02-22 Author has 94 answers
The definition of electric force:
F=kq2d2
If you change the information as stated in the problem:
k(q4)2(d2)2=4kq216d2=kq24d2=F4
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Jeffrey Jordon
Answered 2021-09-30 Author has 2047 answers

Given

Distance between two charge is d and repulsive force is F

Solution

Force between two point charges is given by

f=kq1q2d2

k= coulomb's constant

d= distance between two charges

Now new charges are

q14 and q24

and distance between them is d2

F1=k(q14×q24)(d2)2

=kq1q216d24

F1=kq1q24d2

F1=F4

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