   Chapter 15, Problem 8P

Chapter
Section
Textbook Problem

Four point charges are at the corners of a square of side a as shown in Figure P15.8. Determine the magnitude and direction of the resultant electric force on q, with ke, q, and a left in symbolic form. To determine
The magnitude and direction of resultant electric force.

Solution:

The magnitude of resultant electric force is 4.33(q2kea2) .

The direction of resultant electric force is 45ο above the horizontal.

Explanation

The force diagram is given by, In the above diagram,

• F2q is the force due to the charge 2q on q.
• F3q is the force due to the charge 3q on q.

Formula to calculate the force due to the charge 2q on q is,

F2q=2q2kea2      (I)

• ke is the Coulomb constant.

Formula to calculate the force due to the charge 3q on q is,

F3q=3q2ke(2a)2       (II)

Net force along the x direction is,

Fx=F2q+F3qcos45ο       (III)

Net force along the y direction is,

Fy=F2q+F3qsin45ο       (IV)

Formula to calculate the magnitude of resultant electric force is,

FR=Fx2+Fy2       (V)

Substitute Equations (III) and (IV) in (V).

FR=(F2q+F3qcos45ο)2+(F2q+F3qsin45ο)2      (VI)

Substitute Equations (I) and (II) in (VI).

FR=(2q2kea2+3q2ke(2a)2cos45ο)2+(2q2kea2+3q2ke(2a)2sin45ο)2=(q2kea2)(2+32cos45ο)2+(2+32sin45ο)2=4.33(q2kea2)

Formula to calculate the direction is,

θ=tan1(FyFx)

From Equations (I), (II), (III) and (IV) of section 1,

θ=tan1(2q2kea2+3q2ke(2a)2sin45ο2q2kea2+3q2ke(2a)2cos45ο)=tan1(1)=45ο

Conclusion:

The magnitude of resultant electric force is 4.33(q2kea2) .

The direction of resultant electric force is 45° above the horizontal

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