Electric fields created by charged particles

[perfection256]

Homework Statement

At a particular moment, three small charged balls, one negative and two positive,are located as shown in Figure 13.46. Q1 = 1 nC, Q2 = 5 nC, and Q3 = -2 nC.

Remember that you must first convert all quantities to S.I. units. 1 nC = 1 nanocoulomb = 1e-9 C.

(a1) What is the electric field at the location of Q1, due to Q2?
(a2) What is the electric field at the location of Q1, due to Q3?


(c1) What is the electric field at location A, due to Q1?
(c2) What is the electric field at location A, due to Q2?
(c3) What is the electric field at location A, due to Q3?

(d) An alpha particle (He2+, containing two protons and two neutrons) is released from rest at location A. At the instant the particle is released, what is the electric force on the alpha particle, due to Q1, Q2 and Q3?

Homework Equations

|E|= kq/|r^2|

The Attempt at a Solution

(a1) (9e^9)(5x10^-9)/(.0016) *<0,.04,0>= <0,1125,0>
(a2) (9e^9)(-2x10^-9)/(.0009) * <.03,0,0>=<300,0,0>

(c1) (9e^9)(1x10^-9)/(.0009)* <.03,0,0>=<300,0,0>
(c2) (9e^9)(5x 10^-9)/(.0025)* <.03,.04,0>=<540,720,0>
(c3) (9e^9)(-2x 10^-9)/(.0016)* <0,.04,0)=<0,-450,0>

 

[__JR__]

The equation you need here is $$E = \frac{kq}{r^2} * \hat{r}$$, where k is the permitivity of free space.

$$\hat{r}$$ represents the radial unit vector from the point charge to the point where you are calculating the electric field, and shouldn't have a magnitude. (I think your * < x,y,z> is indicating that you're multiplying by a magnitude, which would be incorrect).

Also, since this is a vector field, use their coordinate system and stay consistent.

 

[thomate1]

(d) The electric force on the alpha particle would be the sum of the individual electric forces from Q1, Q2, and Q3. Since the alpha particle has a positive charge, it would experience a repulsive force from Q1 and Q2, and an attractive force from Q3. The magnitude of the total electric force can be calculated using the formula F = qE, where q is the charge of the alpha particle and E is the total electric field at location A. The direction of the force would depend on the direction of the individual electric fields at location A.