Question

Two charges are placed on the x axis. One of the charges (q₁ = +7.6 µC) is at x₁ = +3.3 cm and the other (q₂ = -18 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.)


(b) Find the net electric field (magnitude and direction) at x = +6.4 cm. (Use the sign of your answer to indicate the direction along the x-axis.)

Answer 1

Answer :-

(a) The net electric field (magnitude and direction) at x = 0 cm.

• Magnitude :
• Direction     : Along negative X-axis.

(b) The net electric field (magnitude and direction) at x = 6.4 cm.

• Magnitude :
• Direction   : Along positive X-axis.

Explanation :-

Given:-

• 
• 
• 
• 

Let,

• : be the electric produced by at
• : be the electric produced by at
• : be the electric produced by at
• : be the electric produced by at

By Coulomb's we can have expression of the electric field as,

Where,

• : is the proportionality constant.
• 
• : is the distance

(a) The net electric field (magnitude and direction) at x = 0 cm.

1) Magnitude of the electric field due to charge q₁ at x = 0m is given by,

This is the magnitude of electric field due to charge q₁ at x = 0m .

Since the charge q₁ is positive, the direction of this electric field is along negative X-axis. Therefore,

2) Magnitude of the electric field due to charge q₂ at x = 0m is given by,

This is the magnitude of electric field due to charge q₂ at x = 0m .

Since the charge q₂ is negative, the direction of this electric field is along positive X-axis. Therefore,

Therefore, the magnitude and direction of the net electric field at point x = 0m due to charges q₁ and q₂ is given as,

This indicates that the direction of the net electric field is along negative X-axis (left).

Thus the magnitude of the net electric field is ,

(b) The net electric field (magnitude and direction) at x = 6.4 cm.

1) Magnitude of the electric field due to charge q₁ at x = 0.064 m is given by,

This is the magnitude of electric field due to charge q₁ at x = 6.4 cm .

Since the charge q₁ is positive, the direction of this electric field is along positive X-axis. Therefore,

2) Magnitude of the electric field due to charge q₂ at x = 0.064 m is given by,

This is the magnitude of electric field due to charge q₂ at x = 6.4 cm .

Since the charge q₂ is negative, the direction of this electric field is along positive X-axis. Therefore,

Therefore, the magnitude and direction of the net electric field at point x = 6.4 cm due to charges q₁ and q₂ is given as,

This indicates that the direction of the net electric field is along positive X-axis (right).

Thus the magnitude of the net electric field is ,