# Electric fields created by charged particles

• perfection256
So, for the first part,(a1) What is the electric field at the location of Q1, due to Q2?<0, 1125, 0> N/C
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?

|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>

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.

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(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.

## 1. What is an electric field?

An electric field is a physical field that is created by charged particles. It is a region in which a force is exerted on other charged particles, causing them to either attract or repel each other.

## 2. How are electric fields created by charged particles?

Electric fields are created when charged particles, such as protons or electrons, are present. These charged particles create a force that interacts with other charged particles in the surrounding area, resulting in an electric field.

## 3. How does distance affect the strength of an electric field?

The strength of an electric field decreases as the distance from the charged particles increases. This is because the force between charged particles decreases with distance, thus reducing the strength of the electric field.

## 4. What is the relationship between electric fields and electric charges?

Electric fields and electric charges are closely related. Electric charges create electric fields and are affected by the strength of the electric field. The direction of the electric field is determined by the direction of the electric charge.

## 5. Can electric fields be shielded?

Yes, electric fields can be shielded by using conductive materials, such as metal, which can absorb or deflect the electric field. This is why many electronic devices have metal casings, to protect against outside electric fields.

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