particle 1 charge with charge q1=8x10-6 coulombs is located at coordinates x1=(-

Question

particle 1 charge with charge q1=8x10-6 coulombs is located at coordinates x1=(-3,0)meters and particle 2 with charge q2=-14x10-6 coulombs located at coordinates x2=(1,0)meters.

1. if you assume the potential vanishes at infinity, the potential will be zero at some point x0 with -3>x>1 meters. write an expression for x0, in terms of q1,q2,x1, and x2.

2.solve for the numerical value of x0.

3.write the expression for E0 at point x0 in terms of q1,q2,x1 and x2.

4.solve for the numerical value of E0, leave in terms of N/C units.

Explanation / Answer

1)

V1 + V2 = 0

==> k q1/r1 = k q2/r2

==> k q1/(3-x) = k q2/(1+x)

==> q1/(3-x) = q2/(1+x)

==> (1+x)/(3-x) = q2/q1

==> 1 + x = 3 * (q2/q1) - x * (q2/q1)

==> x * (1 + q2/q1) = 3 (q2/q1) - 1

==> x * ((q1+q2)/q1) = (3 q2 - q1)/q1

==> x0 = -(3 q2 - q1)/(q1+q2)

2)

==> x0 = -(3 q2 - q1)/(q1+q2)

==> x0 = -(3*14 - 8)/(8+14)

==> x0 = -1.55 m

3)

E0 = k q1/r1^2 + k q2/r2^2

==> E0 = k q1/(3 - x0)^2 + k q2/(1 + x0)^2

4)

E0 = k q1/(3-x)^2 + k q2/(1+x)^2

==> E0 = 9e9*8e-6/(3-1.5455)y2 + 9e9*14e-6/(1+1.5455)y2

==> E0 = 5.4 * 10^4 N/C

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