A test charge of 5.0x10^-6 C is moved 2.0 cm through an electric force of 6.0 x 10^-7 N .
What is the change in evergy of the test charge?
What is the change in voltage on the charge?
Help
I need to know the signs. If you are pushing the test charge up then the potential energy goes up.
Work = force * distance = increase in energy
= 6*10^-7 * 2* 10^-2 = 12 * 10^-9 = 1.2 * 10^-8 Joules
Voltage change = energy change / charge
= 1.2 * 10^-8 / 5 * 10*-6 = .24 *10^-2
= 2.4 * 10^-3 volts
Hello damon... I'm not sure what happened there... what happened to the 5.0 x 10^-6 Could you show me again, as I don't quite understand it.
thanks
martin
You did not need the 5*10^-6 for the first part of the question. You were given the force and work = change in energy = force times distance
You do need it in the second part where you divide the energy by the charge to get voltage (which is potential energy per unit charge)
1.2 / 5 = .24 and 10^-8/10^-6 = 10^-2
To find the change in energy of the test charge, you can use the equation:
ΔE = F * d
where ΔE represents the change in energy, F is the electric force exerted on the charge, and d is the displacement of the charge.
Plugging in the given values:
F = 6.0 x 10^-7 N
d = 2.0 cm = 0.02 m
ΔE = (6.0 x 10^-7 N) * (0.02 m)
Now, let's calculate the result:
ΔE = 1.2 x 10^-8 J
Therefore, the change in energy of the test charge is 1.2 x 10^-8 J.
To determine the change in voltage on the charge, you can use the equation:
ΔV = ΔE / q
where ΔV represents the change in voltage, ΔE is the change in energy of the charge, and q is the charge itself.
Plugging in the given values:
ΔE = 1.2 x 10^-8 J
q = 5.0 x 10^-6 C
ΔV = (1.2 x 10^-8 J) / (5.0 x 10^-6 C)
Now, let's calculate the result:
ΔV = 2.4 V
Therefore, the change in voltage on the charge is 2.4 V.