Question

Potential energy of two equal positive charges 1 micro coulomb each held 1m apart in air is

Answer 1

Final Answer :
$9 \times {10}^{ - 3} \: joules$

Steps and Understanding :
1) We know that,
Electrostatic Potential Energy of system of two point charges is given by

kq(1)q(2) / r
where q(1),q(2) are two point charges.
r = separation between them.
Here,
q(1) = q(2) = 1 * 10^(-6) C
r = 1m

2) Then,
U = $\frac{9 \times {10}^{ 9} \times {10}^{ - 6} \times {10}^{ - 6} }{ 1 } \\ = > 9 \times {10}^{ - 3} joules$

Therefore, Potential Energy required is 9 * 10^(-3) J

Answer 2

Given conditions ⇒

Magnitude of the Charge = q₁ = q₂ = 1 μC
 = 10⁻⁶ C.

Distance between the Charges(r) = 1 m.
Coulomb constant = 9 × 10⁹ N/C²m²

Using the Formula,

  U = kq²/r
⇒ U = 9 × 10⁹ × (10⁻⁶)²/1
∴ U = 9 × 10⁹ × 10⁻¹²
∴ U = 9 × 10⁻³ J.
∴ U = 9 mJ.


Hence, the Potential energy of the Equal positive charges of 1 micro coulomb each held 1 m apart in air is 9 mJ. 



Hope it helps.