Point charges Q 1 = + 3·10 -6 C and Q 2 = -6·10 -6 C are fixed on the x axis at

Question

Point charges Q1 = + 3·10-6 C and Q2 = -6·10-6 C are fixed on the x axis at x = -8 m and x = +8 m, respectively.

a.) Sketch the electric potential (voltage) on the x axis for this system.

b.) Your sketch should show that there is a point on the x axis between the two charges where V=0.
Is this point closer to Q1 or Q2? I already got it was Q1.

Looking at your graph, is there an electric field at this point? Yes; No; Not Enough Info

c.) Find the x coordinate of the point referred to in part b.

d.) A +2 millicoulomb charge placed at rest at this point would have:
a potential energy of:  J (Anything significant about this value?)
a kinetic energy of:  J
and it will move: in the +x-dir; in the +y-dir; it will not move since PE=0; not enough info

e.) When this charge passes through the origin, it will have:
a potential energy of:
a kinetic energy of:

Explanation / Answer

b)

The point for zero potential lie at origin since origin is at same distance from two equal and opposite charges. hence the electric Potential by these two charges being equal and opposite cancel out

at the origin, there will be elecric field since electric field between the opposite charges add up in same direction

c)

(0,0)

d)

PE = 0    Since electric potential is 0 at Origin

KE = 0 since the charge is at rest

There is net electric field at origin , hence the charge will move in the direction of electric field which is Positive X-direction.

e)

PE = 0

KE = 0

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