Two particles having charges of +0.500nC and +8.00nC are separated by a distance of 1.20m. (a) at what point along the line connecting the two charges is the net electric field due to the two charges equal to zero? (b) where would the net electric field be zero if one of the charges were negative?

To find the point along the line connecting the two charges where the net electric field is zero, we can use the principle of superposition. The net electric field at any point is the vector sum of the electric fields due to the individual charges.

Let's assume the positive charge of +0.500nC is at point A, and the positive charge of +8.00nC is at point B, and we need to find the point C along the line AB where the net electric field is zero.

(a) To find the point C, we can start by calculating the electric fields due to each charge individually at point C and then add them vectorially to get zero net electric field.

1. Calculate the electric field due to the charge at point A:
The electric field due to a point charge is given by the equation:
Electric field (E) = k * (Q / r^2)
where:
- k is the electrostatic constant (k ≈ 9 x 10^9 Nm^2/C^2)
- Q is the charge (in coulombs)
- r is the distance between the charge and the point where the electric field is being calculated (in meters)

Electric field at point C due to the +0.500nC charge at point A:
E_A = k * (0.500nC) / (AC^2)

2. Calculate the electric field due to the charge at point B:
Similarly, the electric field at point C due to the +8.00nC charge at point B can be calculated using the same formula as above:
E_B = k * (8.00nC) / (BC^2)

3. Calculate the net electric field at point C:
The net electric field at point C is the vector sum of the electric fields due to the charges at A and B:
E_net = E_A + E_B

You can now solve for point C by finding the distance along the line AB where the net electric field, E_net, becomes zero.

(b) If one of the charges were negative, the direction of the electric field contribution would change. In that case, the point along the line connecting the charges where the net electric field is zero would also change. To find this point, you would follow the same steps as in part (a), but considering the charge at B as negative. The algebraic sign change would affect the vector sum of the electric fields, resulting in a different point where the net electric field is zero.

To find the point along the line connecting the two charges where the net electric field is zero, we can use the principle of superposition.

(a) Let's assume that the positive charge of +0.500 nC is located at the origin O, and the positive charge of +8.00 nC is located at point P, which is 1.20 m to the right of the origin.

We want to find the point where the net electric field is zero. At this point, the electric field due to the +0.500 nC charge is equal in magnitude but opposite in direction to the electric field due to the +8.00 nC charge.

Let's consider a point R on the line OP, which is at a distance x from the origin O.

The electric field, E1, due to the +0.500 nC charge at point R is given by Coulomb's Law:

E1 = k * q1 / r1^2

where k is the electrostatic constant (k = 9.0 × 10^9 N m^2/C^2), q1 is the charge (+0.500 nC), and r1 is the distance from the charge (+0.500 nC) to point R.

Similarly, the electric field, E2, due to the +8.00 nC charge at point R is given by:

E2 = k * q2 / r2^2

where q2 is the charge (+8.00 nC), and r2 is the distance from the charge (+8.00 nC) to point R.

At the point where the net electric field is zero, E1 should be equal in magnitude but opposite in direction to E2. Mathematically, this can be written as:

E1 = -E2

Therefore,

k * q1 / r1^2 = -k * q2 / r2^2

Since r1 = x and r2 = 1.20 - x (as point R is at a distance x from O and (1.20 - x) from P), we can rewrite the equation as:

k * q1 / x^2 = -k * q2 / (1.20 - x)^2

Now let's solve this equation for x.

k * q1 / x^2 = k * q2 / (1.20 - x)^2

Simplifying,

q1 / x^2 = q2 / (1.20 - x)^2

Plugging in the values for q1, q2, and k:

(0.500 × 10^-9 C) / x^2 = (8.00 × 10^-9 C) / (1.20 - x)^2

Cross-multiplying,

(0.500 × 10^-9 C) × (1.20 - x)^2 = (8.00 × 10^-9 C) × x^2

Expanding and simplifying,

(1.44 - 2.4x + x^2) = 64x^2

Rearranging,

65x^2 + 2.4x - 1.44 = 0

Now we can solve this quadratic equation to find the value of x.

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