Point charge attractioning a positive charge from infinity to 30cm

[goWlfpack]

Homework Statement

A point charge of 2.0 10-9 C is located at the origin. How much work is required to bring a positive charge of 4.0 10-9 C from infinity to the location x = 30.0 cm?

and it requests the answer in J

Homework Equations

W=-q2(delta)V

DeltaV = Ke q1/d

where delta v is potential difference, W is work, q2 and q1 are the charges, Ke is coulombs constant, and d is distance

The Attempt at a Solution

I plugged in all of the variables using 4.0 *10 ^ -9 as my q1 and 2.0*10^-9 as my q2 and that was wrong so i switched them.. again wrong..

 

[Doc Al]

Switching q1 and q2 won't change anything. Show your final expression for the work done and the numbers you plugged in. Is it positive or negative?

 

[baba89]

I would first clarify the problem by asking for the units of the given charges and the distance. Assuming that the charges are given in coulombs (C) and the distance is given in meters (m), I would then proceed with solving the problem using the equations provided.

First, I would calculate the potential difference (deltaV) between the two charges using the equation deltaV = (Ke*q1)/d. Plugging in the values, I get deltaV = (8.99*10^9 N*m^2/C^2)*(4.0*10^-9 C)/0.30 m = 1.198*10^5 V.

Next, I would use the equation W = -q2*deltaV to calculate the work required to bring the positive charge from infinity to x = 30.0 cm. Plugging in the values, I get W = -(2.0*10^-9 C)*(1.198*10^5 V) = -2.396*10^-4 J.

Therefore, the work required to bring the positive charge from infinity to x = 30.0 cm is 2.396*10^-4 J.