# What is the electric potential of a 2.2 µC charge at a distance of 6.3 m from the charge? Recall that Coulomb’s constant is k = 8.99 × 109 N • ___ V What is the electric potential at a distance of 99 m from the charge?___ V

Answer: 1) The electric potential at a distance r from a single point charge is given by

where k is the Coulomb's constant, q is the charge and r is the distance from the charge.

The charge in this problem is

So the potential at distance  is

2) By using the same formula as before, we can find the electric potential at distance r=99 m from the charge:

3100 V

and

200v

## Related Questions

What is the name of the energy that results from the interaction of charged particles?

Electrostatic energy perhaps ... potential until one charge is released and starts to move.

If a trapezoidal channel has side slopes of 1:1, hydraulic depth is 5 feet, the bottom width is 8 feet, flow is 2,312 cubic feet per seconds and the Manning's "n" is 0.013, what is the slope of the channel? g

Explanation:

given,

side slope = 1 : 1

hydraulic depth(y) = 5 ft

bottom width (b)= 8 ft

x = 1

Q = 2,312 ft³/s

n = 0.013

slope of channel = ?

R = 4.69 m

using manning's equation

S = 0.406

A driver sets out on a journey. for the first half of the distance she drives at the leisurely pace of 30 mi/h; during the second half she drives 60 mi/h. what is her average speed on this trip?

The average speed of a moving object is the rate of change of a certain distance with respect with time. It is equal to the total distance that was traveled by the object over the total time it takes to travel that distance. For this problem we need to assume that the total distance that was traveled would be equal to 120 miles. So, for the first half of the distance or 60 miles at a speed of 30 miles per hour, the time taken would be two hours. For the remaining 60 miles at a speed of 60 miles per hour, 1 hour is total time traveled. So, we calculate the average speed as follows:

Average speed = total distance / total time
Average speed = 120 miles / 2 hr + 1 hr
Average speed = 40 mi / hr

A 77.0-kg ice hockey goalie, originally at rest, catches a 0.125-kg hockey puck slapped at him at a velocity of 37.5m/s. Suppose the goalie and the ice puck have an elastic collision and the puck is reflected back in the direction from which it came.What would the final velocities of the goalie and the puck be in this case? Assume that the collision is completely elastic. A. v goalie = _____ m / s
B. v puck = ______ m / s

given,

Mass of ice hockey goalie = (M) = 77 Kg

mass of pluck = (m) = 0.125 Kg

velocity of pluck (u₁)= 37.5 m/s

u₂ = 0

Let v₁ and v₂ are the velocity of m₁ and m₂

final velocities are

v_1 is velocity of goalie

v_2 is velocity of puck

now,

a) for goalie

b) for pluck