Two charges at the corners of the base of a triangle at a given instance. The ba

Question

Two charges at the corners of the base of a triangle at a given instance. The base of the triangle is horizontal. The left side of trianglg is 4.0 meters long and the right side is 3.0 meters long. The angle at the top apex (the location A) is 90 degrees. The charge on the left is -q1 (q1= 2.0nC) and the charge on the right is q2 (qs is unknown). The magnitude of the total electrical field at A is 1.875 N/C. Gravitational forces are negligible.

A) Find th magnitude (absolute value) q2 of the charge on the right.

B) Find the total electrical potential at A (Take te electrical potentials to be zero infinity).

C) Find the magnitude of the electrical force on -q1, due to q2.

Explanation / Answer

Electric field due to left charge=K*q1/r^2=9*2/4^2=18/16

Electric field due to left charge=K*q1/r^2=9*q2/3^2=q2(in nC)

total field=sqrt[(18/16)^2 +(q2)^2]=1.875

A) ==> [(18/16)^2 +(q2)^2]=(1.875)^2

==> (q2)^2=2.25

==> q2=1.5 nC

B) total electrical potential at A=K*q1/4 +K*q2/3=9*2/4 +9*1.5/3=9 V

C) the electrical force on -q1, due to q2= K*q1*q2/5^2=9 E9 *2*1.5 E-18/5^2=1.08 E-9 N

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