There are 2 point charges. Q1 is 4nC and is at (-2,0). Q2 is -2nC and is at (0,3).
How much work does the electric field do when a third point charge of +3.00 nC is moved from infinitely far away and placed at the origin?
I know infinitely far means Va will be 0. However because there are 2 other charges instead of 1, I do not know how to manipulate k(Q)/d to fit all the charges in. Help!
Energy doesn't care about direction. Just sum the two. kq3q1/r + kq3q2/r.
To find the work done by the electric field, we need to calculate the total electric potential energy before and after the third point charge is moved.
Before the third point charge is moved, the electric potential energy is given by the equation:
U1 = (k * Q1 * q) / r1 + (k * Q2 * q) / r2
where
- U1 is the total electric potential energy,
- k is the electrostatic constant (9 * 10^9 Nm^2/C^2),
- Q1 and Q2 are the charges of the first and second point charges respectively,
- q is the charge of the third point charge, and
- r1 and r2 are the distances between the first and second charges and the third charge respectively.
Since the third point charge is initially placed infinitely far away, we assume the potential energy is zero (Ua = 0). Therefore, we can write the equation as:
0 = (k * Q1 * q) / r1 + (k * Q2 * q) / r2
Next, we move the third point charge to the origin (0, 0). At this new position, the electric potential energy is given by the equation:
Ub = (k * Q1 * q) / r1 + (k * Q2 * q) / r2 + (k * Q3 * q) / r3
where
- Ub is the new total electric potential energy,
- Q3 is the charge of the third point charge, and
- r3 is the distance between the third charge and the charges Q1 and Q2.
Since Q3 is now at the origin, the distances r1 and r2 are the distances between Q3 and Q1, and Q3 and Q2 respectively. These distances can be calculated using the distance formula:
r1 = √((-2 - 0)^2 + (0 - 0)^2)
= √((-2)^2 + (0)^2)
= √(4)
= 2
r2 = √((0 - 0)^2 + (3 - 0)^2)
= √((0)^2 + (3)^2)
= √(9)
= 3
Now, we can substitute the values into the equation to find the new total electric potential energy (Ub). This equation becomes:
Ub = (9 * 10^9) * [(4 * 10^-9 * 3 * 10^-9) / 2 + (-2 * 10^-9 * 3 * 10^-9) / 3]
Simplifying this equation will give you the value of Ub, which represents the new total electric potential energy.
Finally, to find the work done by the electric field, we use the equation:
Work = Ub - Ua
Since Ua is zero, the work done by the electric field is equal to Ub.