For the charge configuration shown in Figure 1 what are the

For the charge configuration shown in (Figure 1), what are the x and y components of the vector sum of the electric forces exerted on particle 3? Suppose that q_1 = -4.00 muC, q_2 = -5.00 muC and + q_3 = +5.00 muC.

Given q1 = -4 x10 ^-6 C

q2 = -5 x10 ^-6 C

q3 = 5 x10 ^-6 C

Coulomb\'s constant K = 8.99 x10 ⁹ Nm ²/C ²

X component of force on q3 due to charge q1 is F = Kq3q1 / 4 ²

Since distance between q3 and q1 along X axis is = 4 m

Substitue values you get , F = (8.99x10 ⁹ )(5 x10 ^-6)(4x10 ^-6)/16
= 11.23 x10 ^-3 N

X component of force on q3 due to charge q2 is F \' = Kq3q1 / 3 ²

Since distance between q3 and q2 along X axis is = 3 m

Substitue values you get , F \' = (8.99x10 ⁹ )(5 x10 ^-6)(5x10 ^-6)/9
= 24.97 x10 ^-3 N

Both F and F \' are in same direction.

Therefore net force on q3 due to q1 and q2 along x direction = F+F \'

= 36.2 x10 ^-3 N

F _x = 36 x10 ^-3 N

-----------------------------------------------------------------------------------------------------------------

Y component of force on q3 due to charge q1 is F = Kq3q1 / 3 ²

Since distance between q3 and q1 along Y axis is = 3 m

Substitue values you get , F = (8.99x10 ⁹ )(5 x10 ^-6)(4x10 ^-6)/9
= 19.96 x10 ^-3 N

Y component of force on q3 due to charge q2 is F \' = Kq3q2 / 4 ²

Since distance between q3 and q2 along X axis is = 3 m

Substitue values you get , F \' = (8.99x10 ⁹ )(5 x10 ^-6)(5x10 ^-6)/16
= 14.04 x10 ^-3 N

Both F and F \' are in opposite in direction.

Therefore net force on q3 due to q1 and q2 along Y direction = F-F \'

= 5.9 x10 ^-3 N along negative Y direction.

F _y = -5.9 x10 ^-3 N

For the charge configuration shown in (Figure 1), what are the x and y components of the vector sum of the electric forces exerted on particle 3? Suppose that

For the charge configuration shown in Figure 1 what are the

Solution

Get Help Now

Submit a Take Down Notice