(Figure 1) shows five electric charges. Four charges with the magnitude of the c

Question

(Figure 1) shows five electric charges. Four charges with the magnitude of the charge 2.0 nC form a square with the size a = 4.0 cm . Positive charge with the magnitude of q = 5.5 nC is placed in the center of the square.

1.) What is the magnitude of the force on the 5.5 nC charge in the middle of the figure due to the four other charges?

2.)What is the direction of the force on the 5.5 nC charge in the middle of the figure due to the four other charges? Give answer as  ?, counterclockwise from the positive x-axis.

Explanation / Answer

Electrostatic force is given by:

F = kq1q2/r^2

Since all four charge on corner are same, so due to them force will be equal in magnitude, which will be

|q1| = 2 nC, & |q2| = q = 5.5 nC

r = sqrt (a^2/4 + a^2/4) = a/sqrt 2

a = 4.0 cm = 0.04 m

So,

|F| = 9*10^9*2*10^-9*5.5*10^-9/(0.04/sqrt 2)^2

|F| = 1/24*10^4 N

Now let's talk about direction, assuming q is at origin, then

Due to charge in first quadrant, since both charge are positive So force on q will be repulsive, towards 45 deg below -ve x-axis in third quadrant.

Due to charge in 2nd quadrant, since both charge are of opposite signs, So force on q will be attractive, towards 45 deg above -ve x-axis in 2ndquadrantt.

Due to charge in 3rd quadrant, since both charge are of opposite signs, So force on q will be attractive, towards 45 deg below -ve x-axis in third quadrant.

Due to charge in 4th quadrant, since both charges are positive So the force on q will be repulsive, towards 45 deg above -ve x-axis in third quadrant.

Now F1 and F3 are below -ve x-axis at 45 deg and F2 and F4 are above -ve axis at 45 deg

F1x = -F1*cos 45 deg & F1y = -F1*sin 45 deg

F2x = -F2*cos 45 deg & F2y = F2*sin 45 deg

F3x = -F3*cos 45 deg & F3y = -F3*sin 45 deg

F4x = -F4*cos 45 deg & F4y = F4*sin 45 deg

Since all forces are equal in magnitude, F1 = F2 = F3 = F4 = F

Now add all of them

Fnet = 4*F*cos 45 deg i + 0 j

Sum of all force in y-direction will be zero.

So

Fnet = 4*1.24*10^-4*cos 45 deg = 3.51*10^-4 N

Direction = -ve x-axis = 180 deg Counter clockwise from +ve x-axis

Please Upvote.

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.