Help, I know the answer is 4.8 X 10^-19, but I can not figure out how to do it or what formula to use. The question is "A charge moves a distance of 1.7 cm in the direction of a uniform electric field whose magnitude is 220 N/C. As the charge moves, its electrical potential energy decreases by 1.8 x 10^-18 J. Find the charge on the moving particle. Thanks.

To find the charge on the moving particle, we can use the equation:

ΔPE = q * ΔV

Where:
- ΔPE is the change in electrical potential energy (given as -1.8 x 10^-18 J in the question)
- q is the charge on the moving particle (what we need to find)
- ΔV is the change in electric potential (related to the electric field and the distance the charge moves)

First, let's find the change in electric potential (ΔV):

ΔV = E * d

Where:
- ΔV is the change in electric potential
- E is the magnitude of the electric field (given as 220 N/C in the question)
- d is the distance the charge moves (given as 1.7 cm, but we need to convert it to meters)

Converting 1.7 cm to meters:
1.7 cm * (1 m / 100 cm) = 0.017 m

Now, substitute the values into the equation to find ΔV:

ΔV = (220 N/C) * (0.017 m) = 3.74 V

Now that we have found ΔV, we can substitute the values into the first equation to find the charge (q):

ΔPE = q * ΔV
-1.8 x 10^-18 J = q * 3.74 V

To isolate q, divide both sides of the equation by 3.74 V:

q = (-1.8 x 10^-18 J) / (3.74 V) = -4.82 x 10^-19 C

Remember that charge is a scalar quantity and can be either positive or negative. In this case, the negative sign indicates that the charge on the moving particle is negative.

Therefore, the charge on the moving particle is approximately -4.82 x 10^-19 C.