Verified answer

Answer:

(i) -kq²/L

(ii) 2kQ/(√3L)

Explanation:

[tex]dU=Fdx[/tex]

[tex]U=\int\limits^\infty_r {F}\,dx=\int\limits^\infty_r {\frac{kq_1q_2}{x^2}} \, dx=kq_1q_2\int\limits^\infty_r {\frac{1}{x^2}} \, dx[/tex]

[tex]=kq_1q_2[\frac{-1}{x}]^{\infty}_r=kq_1q_2(\frac{-1}{\infty}-\frac{-1}{r})=\frac{kq_1q_2}{r}[/tex]

(i) Electrostatic potential energy = kq₁q₂/r

P.E. between Q and q = kQq/L
P.E. between Q and -q = -kQq/L
P.E. between q and -q = -kq²/L

Potential energy of system = kQq/L - kQq/L - kq²/L
[tex]=\frac{-kq^2}{L}[/tex]

(ii) Electrostatic potential = Potential energy per charge = kq/r

Altitude of equilateral triangle = √{L² - (L/2)²} = √3L/2

Potential due to q = kq/(L/2) = 2kq/L
Potential due to -q = -kq/(L/2) = -2kq/L
Potential due to Q = kQ/(√3L/2) = 2kQ/(√3L)

Total potential = 2kq/L - 2kq/L + 2kQ/(√3L)
[tex]=\frac{2kQ}{\sqrt{3}L}[/tex]

Hope this helped :-)