Question

two insulated copper spheres of charge +10 microcoulomb ,-20 microcoulomb have their centres seperated by a distance of 10 cm .what is the force of attraction if they are placed
a)in free space
b) in water of dielectric constant €
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Answer 1

Given:

2 insulated copper spheres of charge +10 microcoulomb and -20 microcoulomb have their centres separated by a distance of 10 cm.

To find:

Electrostatic Force of Attraction

• in free space
• in water of dielectric constant k

Calculation:

In free space:

Let force of attraction be F1

$F1 = \dfrac{1}{4\pi\epsilon_{0}} \bigg \{\dfrac{(10 \times {10}^{ - 6} ) \times (20 \times {10}^{ - 6} )}{ {(0.1)}^{2} } \bigg \}$

$= > F1 = \dfrac{1}{4\pi\epsilon_{0}} \bigg \{\dfrac{(10 \times {10}^{ - 6} ) \times (20 \times {10}^{ - 6} )}{ {10}^{ - 2} } \bigg \}$

$= > F1 = \dfrac{1}{4\pi\epsilon_{0}} \bigg \{\dfrac{(200 \times {10}^{ - 12} ) }{ {10}^{ - 2} } \bigg \}$

$= > F1 = \dfrac{1}{4\pi\epsilon_{0}} \bigg \{200 \times {10}^{ - 10} \bigg \}$

$= > F1 = \dfrac{1}{4\pi\epsilon_{0}} \bigg \{2 \times {10}^{ - 8} \bigg \}$

$= > F1 = 9 \times {10}^{9} \times \bigg \{2 \times {10}^{ - 8} \bigg \}$

$= > F1 = 18 \times 10$

$= > F1 = 180 \: N$

In water of dielectric constant k:

Let force of attraction in water be F2:

$F2= \dfrac{1}{4\pi\epsilon_{0}k} \bigg \{\dfrac{(10 \times {10}^{ - 6} ) \times (20 \times {10}^{ - 6} )}{ {(0.1)}^{2} } \bigg \}$

$= > F2 = \dfrac{1}{4\pi\epsilon_{0}k} \bigg \{\dfrac{(10 \times {10}^{ - 6} ) \times (20 \times {10}^{ - 6} )}{ {10}^{ - 2} } \bigg \}$

$= > F2= \dfrac{1}{4\pi\epsilon_{0}k} \bigg \{\dfrac{(200 \times {10}^{ - 12} ) }{ {10}^{ - 2} } \bigg \}$

$= > F2 = \dfrac{1}{4\pi\epsilon_{0}k} \bigg \{2 \times {10}^{ - 8} \bigg \}$

$= > F2 = \dfrac{9 \times {10}^{9}}{k} \times \bigg \{2 \times {10}^{ - 8} \bigg \}$

$= > F2 = \dfrac{18 \times 10 }{k}$

$= > F2 = \dfrac{180}{k} \: N$

So final answer :

Force of attraction :

• in free space is 180 N
• in water is 180/k N