How does the electric force between two charged objects change when the

Physics

1 answer:

castortr0y [4]3 years ago

3 0

Answer:

When the electric force between two charged objects change, the distance between the objects increases in Iicreasing the separation distance between objects and decreases the force of attraction or repulsion between objects.

Hope it helps.

You might be interested in

I NEED THIS QUICKLY

Yanka [14]

True.

I think that’s the answer.

8 0

4 years ago

A wheelchair ramp is 5.2 m long and 0.8 m high. Calculate the ramp’s mechanical advantage

Arlecino [84]

Answer:

Explanation:

3 0

3 years ago

Equipotential surfaces a) make an angle of 45 degrees with the electric field. b) are parallel to the electric field. c) are per

Schach [20]

Answer:

Option c) are perpendicular to the electric field

Explanation:

Equipotential surfaces are perpendicular to the electric field. the electric field lines are projected outwards from the equipotential surface, i.e., the lines of the electric field are at 90 $^{\circ}$ to the equipotential surface.

Equipotential surface are those surfaces that have the same potential at any point on the surface. Thus the potential difference at any point on the surface is zero due to same potential.

Any charge particle on this surface will move in a perpendicular direction to the Coulombian force. No work is done by the force on a particle moving on an equipotential surface.

7 0

3 years ago

A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo

-BARSIC- [3]

Answer:

It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$ .

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$ .

Note that $\ln 9 = 2\,\ln 3$ , and that $\ln 8 = 3\, \ln 2$ .

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$ , and the initial charge of the capacitor be $q_0$ . Then at time $t$ , the charge stored in the capacitor would be: