Two point charges are separated by 25.0cm (see the figure( Figure 1 ) ). Assume

Question

Two point charges are separated by 25.0cm (see the figure(Figure 1) ). Assume that q1 = 6.30nC and q2 = 13.0nC .

Part A

Find the magnitude of the net electric field these charges produce at point A.

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Part B

Find the direction of the net electric field these charges produce at point A.

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Part C

Find the magnitude of the net electric field these charges produce at point B.

6620

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Part D

Find the direction of the net electric field these charges produce at point B.

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Part E

What would be the magnitude of the electric force this combination of charges would produce on a proton at A?

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Part F

What would be the direction of the electric force this combination of charges would produce on a proton at A?

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Figure 1 of 1

Two point charges are separated by 25.0cm (see the figure(Figure 1) ). Assume that q1 = 6.30nC and q2 = 13.0nC .

Part A

Find the magnitude of the net electric field these charges produce at point A.

E =   N/C  

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Part B

Find the direction of the net electric field these charges produce at point A.

The field is directed to the right. The field is directed to the left.

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Correct

Part C

Find the magnitude of the net electric field these charges produce at point B.

E =

6620

  N/C  

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Correct

Part D

Find the direction of the net electric field these charges produce at point B.

The field is directed to the right. The field is directed to the left.

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Correct

Part E

What would be the magnitude of the electric force this combination of charges would produce on a proton at A?

F =   N  

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Part F

What would be the direction of the electric force this combination of charges would produce on a proton at A?

The force would be directed to the right. The force would be directed to the left.

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Figure 1 of 1

Explanation / Answer

PART-A :

q1 = -6.3*10-9 C ;

q2 = -13*10-9 C ;

d1- Distance of point A from q1 ; d1 = 15*10-2 m.

d2 - Distance of point A from q2 ; d2 = 10*10-2 m.

#Electric field at point A due to q1 , Ea1 = k*q1/d12 = 2520 N/C.

#Electric field at point A due to q2 , Ea2 = k*q2/d22 = 11700 N/C.

- Both fields are in opposite direction.

----Net electric field , Ea = Ea2 - Ea1 = 9180 N/C.

PART-B :

Direction of Ea1 = (-i) that is negative x-axis.

Direction of Ea2 = (+i) that is positive x-axis.

Angle between them = 180 degrees.

Therefore, direction of Ea = Direction of field with higher magnitude = (i) that is positive x-axis

# The field is directed to the right.

PART-C :

d1 - Distance of point B from q1 ; d1 = 10*10-2 m.

d2 - Distance of point B from q2 ; d2 = 35*10-2 m.

#Electric field at point B due to q1 , Eb1 = k*q1/d12 = 5670 N/C.

#Electric field at point B due to q2 , Eb2 = k*q2/d22 = 955 N/C.

- Both fields are in same direction.

--- net electric field at point B , Eb = Eb1 + Eb2 = 6625 N/C.

PART-D :

Both fields are in same direction that is positive x-axis direction.

Therefore, the net electric field will be in their direction.

# The field is directed to the right.

PART-E :

Charge of proton, e = 1.6*10-19 C.

#Force between q1 and e , F1 = k*q1*e/d12 = 4032*10-19 N.

#Force between q2 and e , F2 = k*q2*e/d22 = 18720*10-19 N.

--- net force , F = F2 - F1 = 14688*10-19 N.

PART-F :

Both forces are in exactly opposite directions (angle = 180 degrees).

Therefore , the net force will be in F2 direction (higher magnitude).

# the net force is directed to the right.

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