Question

Point charge A is on the x-axis at x = -3.00 cm. At x= 1.00 cm on the x-axis its electric field is 2800 N/C. Point charge B is also on the x-axis, at x= 5.00 cm. The absolute magnitude of charge B is twice that of A.

A) Find the magnitude and direction of the total electric field at the origin if both A and B are positive.
B) Find the magnitude and direction of the total electric field at the origin if both A and B are negative.
C) Find the magnitude and direction of the total electric field at the origin if A is positive and B is negative.
D)Find the magnitude and direction of the total electric field at the origin if A is negative and B is positive

Answer 1

Answer:

A) If both charges A and B are positive, the electric field vectors point away from the charges, and we can use the principle of superposition to find the total electric field at the origin. The electric field at the origin due to charge A is:

E_A = k*q_A/r_A^2

where k is the Coulomb constant, q_A is the charge of A, and r_A is the distance between A and the origin. Since A is on the x-axis, r_A is simply the distance between A and the origin:

r_A = |-3.00 cm| = 3.00 cm = 0.03 m

Using the given electric field at x=1.00 cm, we can solve for the charge of A:

2800 N/C = k*q_A/(0.01 m)^2

q_A = (2800 N/C)(0.01 m)^2/k = 2.5210^-8 C

The electric field at the origin due to charge B is:

E_B = k*q_B/r_B^2

where q_B is the charge of B, and r_B is the distance between B and the origin:

r_B = |5.00 cm| = 5.00 cm = 0.05 m

Since the absolute magnitude of charge B is twice that of A, q_B = 2q_A = 5.0410^-8 C. Using the principle of superposition, the total electric field at the origin is the vector sum of E_A and E_B. Since both electric fields are along the x-axis, the total electric field at the origin will also be along the x-axis:

E_total = E_A + E_B = kq_A/r_A^2 + kq_B/r_B^2

E_total = (910^9 Nm^2/C^2)(2.5210^-8 C)/(0.03 m)^2 + (910^9 Nm^2/C^2)(5.0410^-8 C)/(0.05 m)^2

E_total = 1.87*10^6 N/C

The direction of the electric field is to the right (positive x-axis direction), since the positive charge B is farther to the right than the positive charge A.

Therefore, the magnitude of the total electric field at the origin if both A and B are positive is 1.87*10^6 N/C to the right.

B) If both charges A and B are negative, the electric field vectors point towards the charges, and we can use the same method as above to find the total electric field at the origin. The electric field at the origin due to charge A is:

E_A = -k*q_A/r_A^2

where the negative sign indicates that the electric field is directed towards charge A. The electric field at the origin due to charge B is:

E_B = -k*q_B/r_B^2

Using the same values for q_A and q_B as in part A, we get:

E_total = E_A + E_B = -kq_A/r_A^2 - kq_B/r_B^2

E_total = -(910^9 Nm^2/C^2)(2.5210^-8 C)/(0.03 m)^2 - (910^9 Nm^2/C^2)(5.0410^-8 C)/(0.05 m)^2

E_total = -2.23*10^6 N/C

The magnitude of the total electric field at the origin is 2.23*10^6 N/C, and the direction is to the left (negative x-axis direction), since the negative charges A and B are both to the right of the origin.

Therefore, the magnitude of the total electric field at the origin if both A and B
are negative is 2.23*10^6 N/C to the left.

C) When A is positive and B is negative, the electric field vectors produced by each point charge will be in opposite directions. To find the total electric field at the origin, we need to add the electric field vectors produced by each charge.

The magnitude of the electric field due to charge A at the origin is given by the formula:

E = k * Q / r^2

where k is the Coulomb constant (9.0 x 10^9 N m^2 / C^2), Q is the magnitude of the charge, and r is the distance between the charge and the origin.

Using this formula, the magnitude of the electric field due to charge A at the origin is:

E_A = k * Q_A / r_A^2
= (9.0 x 10^9 N m^2 / C^2) * Q_A / (0.03 m)^2
= (9.0 x 10^9 N m^2 / C^2) * (|Q_A|) / (0.03 m)^2
= (3.2 x 10^6) * |Q_A| N/C

where |Q_A| is the absolute magnitude of charge A.

Similarly, the magnitude of the electric field due to charge B at the origin is:

E_B = k * Q_B / r_B^2
= (9.0 x 10^9 N m^2 / C^2) * (-2|Q_A|) / (0.05 m)^2
= - (1.4 x 10^6) * |Q_A| N/C

where |Q_B| is the absolute magnitude of charge B.

The total electric field at the origin is the vector sum of E_A and E_B:

E_total = E_A + E_B

Since E_A is positive and E_B is negative, the direction of the total electric field is towards the negative x-axis.

The magnitude of the total electric field is:

|E_total| = |E_A + E_B|
= |E_A| - |E_B|
= (3.2 x 10^6) * |Q_A| - (1.4 x 10^6) * |Q_A|
= (1.8 x 10^6) * |Q_A| N/C

Therefore, the magnitude of the total electric field at the origin is (1.8 x 10^6) times the magnitude of charge A, and the direction of the electric field is towards the negative x-axis.

Answer 2

We can use the electric field equation to find the electric field at the origin due to each point charge, and then add them vectorially to find the total electric field. The electric field due to a point charge is given by:

Step-by-step explanation:

E = k * q / r^2

where k is Coulomb's constant, q is the magnitude of the point charge, and r is the distance from the charge to the point where we want to find the electric field.

A) Both charges are positive. The electric field at the origin due to charge A is:

E_A = k * q_A / r_A^2

where q_A is the magnitude of charge A, r_A is the distance from A to the origin, and we have:

q_A > 0

r_A = 3.00 cm = 0.03 m

The electric field at the origin due to charge B is:

E_B = k * q_B / r_B^2

where q_B = 2 * q_A is the magnitude of charge B, r_B is the distance from B to the origin, and we have:

q_B > 0

r_B = 5.00 cm = 0.05 m

Using Coulomb's constant k = 9.00 x 10^9 N*m^2/C^2, we get:

E_A = (9.00 x 10^9 Nm^2/C^2) * q_A / r_A^2

E_A = (9.00 x 10^9 Nm^2/C^2) * q_A / (0.03 m)^2

E_A = 1.00 x 10^12 q_A N/C

E_B = (9.00 x 10^9 Nm^2/C^2) * q_B / r_B^2

E_B = (9.00 x 10^9 Nm^2/C^2) * 2q_A / (0.05 m)^2

E_B = 7.20 x 10^11 q_A N/C

The total electric field at the origin is the vector sum of E_A and E_B. Since the charges are on the x-axis and the origin is also on the x-axis, the total electric field will be along the x-axis. Therefore, we only need to add the magnitudes of E_A and E_B to get the total electric field:

E_total = |E_A| + |E_B|

E_total = 1.00 x 10^12 q_A N/C + 7.20 x 10^11 q_A N/C

E_total = 1.72 x 10^12 q_A N/C

Substituting q_A = 2800 N/C * (0.01 m)^2 / (9.00 x 10^9 N*m^2/C^2) = 3.11 x 10^-6 C, we get:

E_total = 5.39 N/C

Therefore, the magnitude of the total electric field at the origin is 5.39 N/C, and it is directed along the positive x-axis.

B) Both charges are negative. The only difference from part A is that the charges now have negative signs, so we have:

q_A < 0

q_B = -2 * |q_A| = -2q_A < 0

Substituting these signs into the equations for E_A and E_B, we get:

E_A = -1.00 x 10^12 |q_A| N/C

E_B = -7.20 x 10^11 |q_A| N/C

The total electric field at

not sure if this solves your answer but hope it somewhat helps.