Recent content by Dillion

Homework Statement A -2.0*10^-3 C charge is 0.2 m away from a -6.0*10^-3 C charge. How much work is must be done on the first charge to move it to a distance of 0.9m? Homework Equations F = qs*qt*k/r^2 W = F * d* cos theta The Attempt at a Solution (-2.0*10^-3)(-6.0*10^-3)(9*10^9)/0.2^2...

Gravitational force. The gravitational force will point straight down which will be (9.8 x 1500) = 14700 N.

1/2mvi^2 + mgh = 1/2mvf^2 + mgh 75000+73500=750vf^2+147000 vf = 1.4 as for what forces cause the car to slow down...i would say friction but this problem says frictionless.

Homework Statement A 1500 kg car is approaching an icy (i.e. frictionless) hill as shown below. The car then runs out of gas traveling at a speed of 10 m/s. a) Will the car make it to the top of the hill? picture is attached Homework Equations vf^2 = vi^2 + 2ax[/B]The Attempt at a Solution...

b) V = qsk/r (9*10^9)(3*10^-6)/2 = 13500 (9*10^9)(3*10^-6)/6 = -4500 (electric potential decreases) -4500+13500 = 9000 ?

b) V = qsk/r (9*10^9)(3*10^-6)/2 = 13500 (9*10^9)(3*10^-6)/6 = 4500 = 18000 I still don't understand why this is incorrect. Everything is in the positive direction

Need more help please

I'm not sure how to continue. Having a positive charge near the origin would increase electric potential and if it was closer to the negative charge electric potential would decrease

Wouldn't the signs all be positive because a positive particle would move to the right away from the origin and towards the negative particle at 8 m

b) V = qsk/r (9*10^9)(3*10^-6)/2 = 13500 (9*10^9)(3*10^-6)/6 = 4500 = 18000 is this how I would do find the electric potential at point 2?

oh yeah sorry qs times k which is 9*10^9 all divided by r

point source charge

Electric Potential = qsk/r ?

What is that formula?

Homework Statement A positive charge of 3 microCouloumbs is at the origin. A negative charge with the same magnitude is 8 m away along the x direction. a) What are the magnitude and direction of the electric field at the points between the two charges, 2, and 4, and 6m from the positive charge...