Question

Two point charge qp =+2uc and qQ =-2uc are placed at a distance of 10 cm in air /vacuum ?

Answer 1

Solution:-

$\underline{\rm \: formula \: }$

$\rm \: f = \dfrac{kq_1q_2}{ {r}^{2} }$

So k depends on medium

For air / free space / vaccum :- k = 9 × 10⁹ Nm²/c²

$\underline{\rm \: given \: }$

$\to \rm \: q_1 = + 2 \mu c$

$\rm \to q_2 = - 2 \mu c$

$\to \rm \: distance \: (r) = 10cm = 0.1m$

where

$\mu \: = 10 {}^{ - 6}$

By applying formula

$\rm \: f = \dfrac{kq_1q_2}{ {r}^{2} }$

$\rm \: f \: = \dfrac{9 \times 10 {}^{9} \times - 2 \times 10 {}^{ - 6} \times 2 \times {10}^{ - 6} }{0.1 \times 01}$

$\rm \: f = \dfrac{ - 36 \times 10 {}^{ - 3} }{0.01} = \dfrac{ - 36 \times {10}^{ - 3} }{10 {}^{ - 2} }$

$\rm \: f = \dfrac{ - 36}{10}$

$\rm \: f \: = - 3.6N$

Answer 2

Answer:

formula

\rm \: f = \dfrac{kq_1q_2}{ {r}^{2} }f=

r

2

kq

1

q

2

So k depends on medium

For air / free space / vaccum :- k = 9 × 10⁹ Nm²/c²

\underline{\rm \: given \: }

given

\to \rm \: q_1 = + 2 \mu c→q

1

=+2μc

\rm \to q_2 = - 2 \mu c→q

2

=−2μc

\to \rm \: distance \: (r) = 10cm = 0.1m→distance(r)=10cm=0.1m

where

\mu \: = 10 {}^{ - 6}μ=10

−6

By applying formula

\rm \: f = \dfrac{kq_1q_2}{ {r}^{2} }f=

r

2

kq

1

q

2

\rm \: f \: = \dfrac{9 \times 10 {}^{9} \times - 2 \times 10 {}^{ - 6} \times 2 \times {10}^{ - 6} }{0.1 \times 01}f=

0.1×01

9×10

9

×−2×10

−6

×2×10

−6

\rm \: f = \dfrac{ - 36 \times 10 {}^{ - 3} }{0.01} = \dfrac{ - 36 \times {10}^{ - 3} }{10 {}^{ - 2} }f=

0.01

−36×10

−3

=

10

−2

−36×10

−3

\rm \: f = \dfrac{ - 36}{10}f=

10

−36

\rm \: f \: = - 3.6Nf=−3.6N