Question

What is the force on the charge located at x = +8.00 cm in Figure 17.40(a) given that q = 5.00 muC and a = 7.50? N The point charges in Figure 17.41 are located at the corners of an equilateral triangle 25.0 cm on a side, where q_b = +10.0 muC and q_c = -5.00 muC. (a) Find the electric field at the location of q_a. Magnitude N/C Direction degree (b) What is the force on q_a, given that q_a = +1.50 nC? Magnitude N Direction degree

Answer 1

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Problem 7)

The expression for the electric force

F = k q1q2 / r²

The net force about charge -aq

F = F12 – F32

calculate the distance

F12 = k q1 q2 / r₁₂²

r₁₂ = 8 -3 = 5 cm = 5 10^-2 m

r₃₂ = 11-8 = 3 cm = 3 10^-2 m

calculate the force

Data

q = 5 μC = 5 10^-6 C

a = 7.50

F12 = k q1 q2 / r₁₂²

F12 = 8.987 10⁹ 5 10^-6 (7.50 5 10^-6 )/ (5 10^-2)²

F12 = 6.74 10² N

F32 = k q3 q2 / r₃₂²

F32 = 8.987 10⁹ 5 10^-6 (7.50 5 10^-6 )/ (3 10^-2)²

F32 = 1.87 10³ N

The net force is

F = 6.74 10² N - 1.87 10³ N

F = -1198 N = -1.20 N

The negative sign indicates that the force is directed to the left

Problem 8

The expression for the electric field is

E = k q / r²

r = 25 cm = 0.25 m

Eb = 8.987 10⁹ 10 10^-6 /0.25²

Eb = 1.44 10⁶ N/C

Ec= 8.987 10⁹ 5 10^-6 / 0.25²

Ec = 7.19 10⁵ N/C

These field are in the line of the triangle we use trigonometry to find components

Cos 30 = Eby/Eb

Eby = Eb Cos 30

Eby = 1.44 10⁶ Cos 30

Eby = 1.247 10⁶ N/C

Sin 30 = Ebx /Eb

Ebx = Eb Sin 30

Ebx = 1.44 10⁶ Sin 30

Ebx = 0.72 10⁶ N/C

Ecy = Ec Cos 30

Ecy = 6.227 10⁵ N/C

Ecx = Ec Sin 30

Ecx = 3.595 10⁵ N/C

The net field is the sum of each axis

Ex = Ebx + Ecx

Ex = 0.72 10⁶ + 3.595 10⁵

Ex =10.795 10⁵ N/C

Ey = Eby – Ecy

Ey = 12.47 10⁵ - 6.227 10⁵

Ey = 6.243 10^{5 N/C}

E = sqrt (Ex² + Ey²)

E = 10⁵ sqrt (10.795² + 6.243² )

E = 12.47 10⁵ N/C

Direction

Tan θ = Ey/Ex

Tan θ = 6.243/10.795

θ = 30

Part b)

F = q E

F = 1.5 10^-9 12.47 10⁵

F = 1.87 10^-3 N

As the charge is positive the direction does not change

θ = 30