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Homework Statement "A rod of 13.1 cm long is uniformly charged and has a total charge of -23.2 micro coulombs. Determine the magnitude of the electric field along the axis of the rod at a point 52.1575 cm from the center of the rod. The Coulomb constant is 8.98755e9 N M^2/C^2. Answer in units...

So , intuitively, two "+" equal charges would represent the lines of the electric field for two equal point masses because two "-" charges would mean that electric field lines would have to show up out of no where and that's not possible. Correct?

Yes? Whats choice "B" trying to state?

Gravitational "lines of force" for two equal point masses? Homework Statement "Sketch the gravitational "lines of forces" for two equal point masses. These are isomorphic to the electrostatic lines of force of A. Two equal "+" point charges. B. No pair of point charges separated by the same...

Why would I need to integrate? It already gives me the total charge for the ring. from what points would I integrate? 0 to 8.1?

Homework Statement A uniformly charged ring of radius 8.1 cm has a total charge of 118 micro Coulombs. The value of the Coulomb constant is 8.98755e9 N M^2/C^2. Find the magnitude of the electric field on the axis of the ring at 1.15 cm from the center of the ring. Answer in units of N/C...

do i just do the pyth. theorem to solve for "r" ( square root of 8.1^2 + 1.15^2)?

Homework Statement A uniformly charged ring of radius 8.1 cm has a total charge of 118 micro Coulombs. The value of the Coulomb constant is 8.98755e9 N M^2/C^2. Find the magnitude of the electric field on the axis of the ring at 1.15 cm from the center of the ring. Answer in units of N/C...

it has something to do with the force of the car and truck. and yes the truck has a heavier mass

So i have the following question(s) in my homework assignment (This question is in regards to a car that is pushing a truck that has broken down in the middle of the road and i have to describe the motion/force): Wouldn't the answer be that the car is pushing with less force than the truck is...