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Answer :
a) -60000V is the electric potential at the point (0.30m, 0).
Calculate the distance of each charge from the point (0.30m, 0).
Distance between 10µC charge at origin and point (0.30m, 0) is 0.30m.
Distance between -20µC charge at (0, 0.40m) and point (0.30m, 0) is √((0.40m)² + (0.30m)²)
= √(0.16 + 0.09)
= √0.25
= 0.50m.
Calculate the potential due to each charge at the point (0.30m, 0) using V = k Q / r.
Potential due to 10µC charge:
V1 = (8.99 x 10⁹ N m²/C²) (10 x 10⁻⁶ C) / 0.30m
= 3 x 10⁵ V
= 300,000V.
Potential due to -20µC charge:
V2 = (8.99 x 10⁹ N m²/C²) (-20 x 10⁻⁶ C) / 0.50m
= -3.6 x 10⁵ V
= -360,000V.
Calculate the net potential by adding the potentials due to both charges.
V_net = 300,000V - 360,000V
= -60,000V.
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