The potential due to the charge Q_{1}:

V_{1} = kq_{1} / r_{1} = 9.0 × 10^{9} N·m^{2}·C^{–2 }× –5.4 × 10^{-6 }C / 2.0 m = –24300 V

The potential due to the charge Q_{2}:

V_{2} = kq_{2} / r_{2} = 9.0 × 10^{9} N·m^{2}·C^{–2 }× 7.6 × 10^{-6 }C / 4.0 m = +17100 V

The net potential is –24300 V + 17100 V = –7200 V

Thank you, but do you know the equation for (B)

To find the intensity of the electric field formed by a charge at some point, use the formula E = kq /r^2

where k = 9 * 10^9 N m^2 / C^2 is the constant, q is the charge, and r is the distance from the charge to the point.

Thank you that helped a lot

You are welcome.

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