Q1: A charge of -2×10 -9 C is at the origin and a charge of 6.2×10 -9 C is on th

Question

Q1: A charge of -2×10-9 C is at the origin and a charge of 6.2×10-9 C is on the x-axis at x = 3m. At what two locations on the x-axis (xpositive, xnegative) is the potential zero?

xpositive =

b)

xnegative =

I got 4.18 and -1.108 but those were not correct. Please Help!

Q2: A positive charge of 2.7×10-9 C is brought in from infinity to a point (2m,0).

How much work did the electric field do?

W = Found the correct answer to be 0J

b) Now another positive charge of 2.7×10-9 C is moved from infinity to the coordinates (1m, 2m).

How much additional work was done by the electric field? (The original charge is held in place.)

W = ??

3)

Finally, a negative charge of -2.7×10-9 C is moved from infinity to the coordinates (-2m,0).

How much additional work was done by the electric field? (The original two charges is held in place.)

W = ??

Please help me with part b and c! Greatly appreciated.

Explanation / Answer

Qns 1 )

V = kq/x

kq1/x + kq2/(3-x)

q1/x = -q2/(3-x)

q1(3-x) = -q2*x

2 x 10^-9 (3-x) = 6.2 x 10^-9 x

6 - 2x = 6.2x

x = 0.7317 m

now for negative x

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

-2x10^9(x+3) = -6.2x10^-9 *x

2x + 6 = 6.2x

x = -1.43 m ( from origin 1.43 )

qns 2 )

part b )

change in pE = work

w = kq1q2/r

w = 9x10^9 x 2.7 x 10^-9 x 2.7 x 10^-9 / sqrt(1^2 + 2^2)

w = 2.93 x 10^-8 J

part c )

w = kq1q3/r13 + kq2q3/r23

here r12 = 4 , r23 = sqrt( 13)

w = -3.46 x 10^-8 J

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