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3 years ago

Two positive charges of 6.0 x 10^-6 C are separated by 0.50 m. Calculate the force of electrical attraction

Physics

1 answer:

SpyIntel [72]3 years ago

3 0

Answer:

Force between two charges is given as 1.296 N

Explanation:

As we know that the two charges will attract or repel each other when they are placed near to each other

Here we know that electrostatic force is given as

$F = \frac{kq_1 q_2}{r^2}$

here we have

$q_1 = q_2 = 6 \times 10^{-6} C$

distance between two charges is given as

r = 0.50 m

so we have

$F = \frac{(9\times 10^9)(6 \times 10^{-6})^2}{0.50^2}$

$F = 1.296 N$

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Answer:

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Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and ' $\epsilon_{0}$ ' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

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Part(b):

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Answer:

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