Question

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)​

Answer 1

Answer:

$E_{net}_M=\sqrt{5259150^2+(-2724948.933)^2}\\ E_{net}_M=5923175.281 V/m$

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

Explanation:

The formula for Electric field is:

$E=\frac{kQ}{r^2}$

where:

k is coulomb constant=$8.99*10^{9} Nm^2/C^2$

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:

$E_1=\frac{8.99*10^{9}*3.5*10^{-6}}{0.1^2} \\E_1=3146500 V/m$

For -7.6 μC charge, Magnitude of Electric Field:

$E_2=\frac{8.99*10^{9}*7.6*10^{-6}}{0.1^2} \\E_2=6832400V/m$

Electric field at third corner:

Diagram is attached (According to the diagram)

$E_{net}=(E_2-E_1cos 60)\hat i-E_1sin60\hat j\\E_{net}=(6832400-3146500cos60)\hat i-3146500sin60\hat j\\E_{net}=5259150\hat i-2724948.933\hat j$

Magnitude is given by:

$E_{net}_M=\sqrt{5259150^2+(-2724948.933)^2}\\ E_{net}_M=5923175.281 V/m$

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