Question

Two chages +6 microcoulomb and -2 microcoulomb are seperated by a distance of 30 cm apart in air.find the electrostatic force between the charges

Answer 1

Answer:

Explanation:

F = K Q1 Q2 /R^2

WHERE THE VALUE OF K = 9×10^9

Q1 = +6

Q2= -2

R = 30 × 10^-2 m

   = 0.3 m

F = [9×10^9 × (+6)×(-2)] / 0.3 ^2

  = - 1.2 ×10^12 N

Answer 2

Answer:

Two charges +6 microcoulomb and -2 microcoulomb are separated by a distance of 30 cm apart in the air, the electrostatic force between the charges is 1.2N

Explanation:

• The electrostatic force between two charges $q_{1}$ and $q_{2}$ separated by a distance $r$ is given by  $F=K\frac{q_{1} q_{2} }{r^{2} }$

        where $K=9*10^{9}$

From the question, we have

the charges $q_{1} = 6*10^{-6}C, q_{2} =-2*10^{-6}C$

the distance between the charges $r=30cm=0.3m$

the electrostatic force is

⇒

$K\frac{q_{1} q_{2} }{r^{2} }=\frac{9*10^{9}*6*10^{-6}*2*10^{-6} }{0.3^{2} }$

⇒ $F=1.2N$