Question

Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2 = +8.9 cm. (a) Find the net electric field (magnitude and direction) at x = 0 cm. (Use the sign of your answer to indicate the direction along the x-axis.) N/C (b) Find the net electric field (magnitude and direction) at x = +5.5 cm. (Use the sign of your answer to indicate the direction along the x-axis.) N/C

Answer 1

To find the net electric field at x = 0 cm, calculate the electric field due to the charges q1 and q2 and add them together. The net electric field at x = 5.5 cm will be directed towards the negative charge. The direction of the net electric field is positive along the x-axis at x = 0 cm and negative along the x-axis at x = 5.5 cm.

Step-by-step explanation:

To find the net electric field at x = 0 cm, we need to calculate the electric field due to both charges q1 and q2 and then add them together. The formula for electric field due to a point charge is given by:

E = k * (|q| / r^2)

Where k is the electrostatic constant (k = 8.99 x 10^9 N m^2/C^2), q is the charge, and r is the distance of the point from the charge.

For q1 at x1 = +3.1 cm:

E1 = (8.99 x 10^9 N m^2/C^2) * (7.7 x 10^-6 C) / (0.031 m)^2

For q2 at x2 = +8.9 cm:

E2 = (8.99 x 10^9 N m^2/C^2) * (-19 x 10^-6 C) / (0.089 m)^2

To find the direction of the net electric field, we need to consider the signs of the charges. Since q1 is positive and q2 is negative, the net electric field at x = 0 cm will be directed towards the positive charge. Hence, the direction of the net electric field is positive along the x-axis.

To find the net electric field at x = 5.5 cm, we follow the same steps as above, but now we calculate the electric field at that point:

E1 = (8.99 x 10^9 N m^2/C^2) * (7.7 x 10^-6 C) / (0.055 m)^2

E2 = (8.99 x 10^9 N m^2/C^2) * (-19 x 10^-6 C) / (0.055 m)^2

Again, considering the signs of the charges, the net electric field at x = 5.5 cm will be directed towards the negative charge. Hence, the direction of the net electric field is negative along the x-axis.

Answer 2

a)
`E=50.53* 10^(6)\ N/C`

The direction will be negative direction.

b)
`E=268.22* 10^(6)\ N/C`

The direction will be positive direction.

Step-by-step explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

`E=K(q)/(r^2)`

`E_1=K(q_1)/(r_1^2)`

`E_2=K(q_2)/(r_2^2)`

Now by putting the va;ues

a)

`E_1=9* 10^9* (7.7* 10^(-6))/(0.031^2)\ N/C`

`E_1=72.11* 10^(6)\ N/C`

`E_2=9* 10^9* (19* 10^(-6))/(0.089^2)\ N/C`

`E_2=21.58* 10^(6)\ N/C`

The net electric field

`E=E_1-E_2`

`E=50.53* 10^(6)\ N/C`

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

b)

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

`E_1=9* 10^9* (7.7* 10^(-6))/(0.024^2)\ N/C`

`E_1=120.3* 10^(6)\ N/C`

`E_2=9* 10^9* (19* 10^(-6))/(0.034^2)\ N/C`

`E_2=147.92* 10^(6)\ N/C`

The net electric field

`E=E_1+E_2`

`E=268.22* 10^(6)\ N/C`

The direction will be positive direction.