Charges of 3.5 µC and −7.6 µC are placed at two corners of an equilateral triangle with sides of 0.1 m. At the third corner, what is the electric field magnitude created by these two charges? (ke = 8.99 × 109 N·m2/C2)

The electric field magnitude created by these two charges=5923175.281 V/m

Explanation:

The formula for Electric field is:

where:

k is coulomb constant=

r is the distance

Q is the charge.

It is a equilateral triangle with all angles 60 degree.

For 3.5μC charge, Magnitude of Electric Field:

For -7.6 μC charge, Magnitude of Electric Field:

Electric field at third corner:

Diagram is attached (According to the diagram)

Magnitude is given by:

The electric field magnitude created by these two charges=5923175.281 V/m

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg