Question

Problem 17.43 Part A Constants Find the magnitude of the electric field this combination of charges produces at point FP Express your answer in newtons per coulomb to three significant figures. Three negative point charges lie along a line as shown in (Figure 1). Point P lies 6.00 cm from the ga charge measured perpendicular to the line connecting the three charges. Assume that q 5.30 pC and g- 2.35 C N/C Submit Request Answer Figure く 1011 Part B 91 8.00 ㎝:6.00 cm Find the direction of the electric field this combination of charges produces at point P. O toward the a charge O toward the bottom gi charge o toward the top q, charge 92 8.00 cm 91

Answer 1

R = AP = distance of charge q_1 from P = squareroot (8^2 + 6^2) = 10 cm = 0.10 m a = BP = distance of charge q_2 from P = 6 cm = 0.06 m q_1 = magnitude of charge = 5.30 x 10^-6 C q_2 = magnitude of charge = 2.35 times 10^-6 C E_1 = electric field by charge q_1at P = k q_1/r^2 = (9 times 10^9) (5.30 times 10^-6)/(0.10)^2 = 4.77 times 10^6 N/C E_2 = electric field by charge q_2 at P = k q_2 /a^2 = (9 times 109) (2.35 times 10^-6)/(0.06)^2 = 5.875 times 106 N/C theta = tan^-1(8/6) = tan^-1(1.33) = 53.06 deg Components E_1 Sin theta along the Y-direction being equal and opposite cancel out . hence the magnitude of net electric field along the X-direction is given as E = 2 E_1 Cos theta + E_2 E = 2 (4.77 times 10^6) Cos 53.06 + (5.875 times 10^6)

r = AP = distance of charge q₁ from P = sqrt(8² + 6²) = 10 cm = 0.10 m

a = BP = distance of charge q₂ from P = 6 cm = 0.06 m

q₁ = magnitude of charge = 5.30 x 10-6 C

q₂ = magnitude of charge = 2.35 x 10^-6 C

E₁ = electric field by charge q₁ at P = k q₁ /r² = (9 x 10⁹) (5.30 x 10^-6)/(0.10)²= 4.77 x 10⁶ N/C

E₂ = electric field by charge q₂ at P = k q₂ /a² = (9 x 10⁹) (2.35 x 10^-6)/(0.06)²= 5.875 x 10⁶ N/C

$\theta$ = tan^-1(8/6) = tan^-1(1.33) = 53.06 deg

Components E₁ Sin$\theta$ along the Y-direction being equal and opposite cancel out . hence the magnitude of net electric field along the X-direction is given as

E = 2 E₁ Cos$\theta$ + E₂

E = 2 (4.77 x 10⁶) Cos53.06 + (5.875 x 10⁶)

E = 1.16 x 10⁷ N/C

Part B)

direction of net electric field comes out to be

towards the charge q₂