Question

three point charges q1(=-1 microCoulomb), q2(=+2 micro Coulomb) and q3(=-1 micro Coulomb) are placed in air in a line.The distance between q1 and q2 is 1m and q2 and q3 is also 1m. Calculate the electrostatic energy of the system.

Answer 1

Answer:

15.75 * 10⁻³ N

Explanation:

Three point charges q1(=-1 microCoulomb), q2(=+2 micro Coulomb) and q3(=-1 micro Coulomb) are placed in air in a line.The distance between q1 and q2 is 1m and q2 and q3 is also 1m. Calculate the electrostatic energy of the system.

q₁   , q₂    q₃

q₁ = -1 μC  

q₂ = 2 μC

q₃ = = -1 μC

Force between q₁ & q₂ attractive  also q₃ & q₂

F₁₂ = k 1 * 10⁻⁶ * 2 * 10⁻⁶ / 1² = 2k * 10⁻¹²

F₂₃ =  k 1 * 10⁻⁶ * 2 * 10⁻⁶ / 1² = 2k * 10⁻¹²

Force on q₂ gets balanced

While

attraction force of 2k   at  q₁ & q₃

Force between q₁ & q₃ will be repulsive

F₁₃ =  k 1 * 10⁻⁶ * 1 * 10⁻⁶ / 2²  = k * 10⁻¹² /4

Net Attractive force between q₁ & q₃  =  2k * 10⁻¹² - k * 10⁻¹² /4

= (7k/4) * 10⁻¹²

k = 9 * 10⁹ Nm²/C²

= 7 * 9 * 10⁹/4 *  10⁻¹²

= 15.75 * 10⁻³ N

Answer 2

The answer is verified.