Answer :
Answer:
512.5 mJ
Explanation:
Let the two identical charges be q = +35 µC and distance between them be r₁ = 46 cm. A charge q' = +0.50 µC located mid-point between them is at r₂ = 46 cm/2 = 23 cm = 0.23 m.
The electric potential at this point due to the two charges q is thus
V = kq/r₂ + kq/r₂
= 2kq/r₂
= 2 × 9 × 10⁹ Nm²/C² × 35 × 10⁻⁶ C/0.23 m
= 630/0.23 × 10³ V
= 2739.13 × 10³ V
= 2.739 MV
When the charge q' is moved 12 cm closer to either of the two charges, its distance from each charge is now r₃ = r₂ + 12 cm = 23 cm + 12 = 35 cm = 0.35 m and r₄ = r₂ - 12 cm = 23 cm - 12 cm = 11 cm = 0.11 cm.
So, the new electric potential at this point is
V' = kq/r₃ + kq/r₄
= kq(1/r₃ + 1/r₄)
= 9 × 10⁹ Nm²/C² × 35 × 10⁻⁶ C(1/0.35 m + 1/0.11 m)
= 315 × 10³(2.857 + 9.091) V
= 315 × 10³ (11.948) V
= 3763.62 × 10³ V
= 3.764 MV
Now, the work done in moving the charge q' to the point 12 cm from either charge is
W = q'(V' - V)
= 0.5 × 10⁻⁶ C(3.764 MV - 2.739 MV)
= 0.5 × 10⁻⁶ C(1.025 × 10⁶) V
= 0.5125 J
= 512.5 mJ
Work done will be: [tex]512.5 \mu J[/tex].
Given:
q = +35 µC
q' = +0.50 µC
r₁ = 46 cm
r₂ = 46 cm/2 = 23 cm = 0.23 m
The electric potential at this point due to the two charges q is thus
[tex]V = \frac{kq}{r_2}+\frac{kq}{r_2}\\\\ V= \frac{2kq}{r_2}\\\\V= \frac{2 * 9 * 10^9Nm^2/C^2 * 35 * 10^{-6} C}{0.23 m}\\\\V= \frac{630}{0.23*10^3}V\\\\V= 2739.13 * 10^3 V\\\\V= 2.739 \mu V[/tex]
When the charge q' is moved 12 cm closer to either of the two charges, its distance from each charge is now;
r₃ = r₂ + 12 cm = 23 cm + 12 = 35 cm = 0.35 m and
r₄ = r₂ - 12 cm = 23 cm - 12 cm = 11 cm = 0.11 cm.
Thus, the new electric potential at this point is
[tex]V' = \frac{kq}{r_3} + \frac{kq}{r_4}\\\\V= kq(\frac{1}{r_3}+\frac{1}{r_4})\\\\V= 9 * 10^9 Nm^2/C^2 * 35 * 10^{-6} C (\frac{1}{0.35m}+\frac{1}{0.11m})\\\\V= 315 * 10^3(2.857 + 9.091) V\\\\V= 315 * 10^3 (11.948) V\\\\V= 3763.62 * 10^3 V\\\\V= 3.764 \mu V[/tex]
Now, the work done in moving the charge q' to the point 12 cm from either charge is:
[tex]W = q'(V' - V)\\\\W= 0.5 * 10^{-6} C(3.764 MV - 2.739 MV)\\\\W= 0.5 *10^{-6} C(1.025 * 10^6) V\\\\W= 0.5125 J\\\\W= 512.5 \mu J[/tex]
Thus, the work done will be: [tex]512.5 \mu J[/tex].
Find out more information on "Work done" here:
brainly.com/question/25573309