Question

find the force of interaction between 60stat coulomb and 3.75stat coulomb spaced 7.5cm apart in transformer oil (€r=2.2) in 10^-4N​

a. 8.15
b. 5.18
c. 1.518
d. 1.815

Answer 1

Answer:

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Answer 2

Given: 60 stat coulomb charge and 3.75 stat coulomb charge are present at distance 7.5 cm apart in transformer oil (€r= 2.2)

To find: Force of attraction in terms of 10^-4 N

Explanation: Distance (d)= 7.5 cm = 7.5 * 10^-2 m

60 stat coulomb(q1)= 60 * 3.33* 10^-10 Coulomb

= 200* 10^-10 Coulomb

3.75 stat coulomb(q2)= 3.75 * 3.33* 10^-10 Coulomb

= 12.5 * 10^-10 Coulomb

$f = \frac{kq1q2}{€r \: {d}^{2} }$

where k = 9* 10^9

$f = \frac{9 \times {10}^{9} \times 200 \times {10}^{ - 10} \times 12.5 \times {10}^{ - 10} }{2.2 \times ( {7.5 \times {10}^{ - 2}) }^{2} }$

=181.8 * 10^ -7 N

= 0.1818* 10^-4 N

Therefore, the force of attraction in the given case is 0.1818* 10^-4 N.