Electric Field Video Lessons

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Problem: In the figure, the point charges are located at the corners of an equilateral triangle 21 cm on a side. Part (a) Find the magnitude of the electric field in N/C at the location of qa given that qb=10.8 uC and qc=4.8 uC. Part (b) Find the direction of the electric field at qa, in degrees above the negative x-axis with origin at qaPart (c) What is the magnitude of the force in N on qa given that qa=1.9 nC?Part (d) What is the direction of the force on qa, in degrees above the negative x-axis with origin at qa?

FREE Expert Solution

Electric field:

E=kqr2

Part (a)

The electric field at qa due to qb is:

Eab=(9×109)(10.8×10-6)(21×10-2)2

Eab = 2.204 × 106 N/C

The electric field at qa due to qc is:

Eac=(9×109)(4.8×10-6)(21×10-2)2

Eac = 9.80 × 105 N/C

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Problem Details

In the figure, the point charges are located at the corners of an equilateral triangle 21 cm on a side.

 Part (a) Find the magnitude of the electric field in N/C at the location of qa given that qb=10.8 uC and qc=4.8 uC. 

Part (b) Find the direction of the electric field at qa, in degrees above the negative x-axis with origin at qa

Part (c) What is the magnitude of the force in N on qa given that qa=1.9 nC?

Part (d) What is the direction of the force on qa, in degrees above the negative x-axis with origin at qa?

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