Question

The table below lists the electric field at various locations along the +-axis. The negative sign means that the electric field points in the -x-direction. Electric Field, E (kN/C) -295.0 Position, r (m) 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 188.8 156.1 96.3 83.9 (a) Assuming that the field was created by a single point charge, determine the value and location of this charge QNumber Units rNumber Units (b) What is the electric field for the two values that are missing in the table? E( 3.00 m)- Number Units E (z-6.00 m)-| Number Units

Answer 1

a)

let the charge is placed at distance "a" from the origin

E₂ = electric field by the charge at x = 2 , = 295 x 10³ N/C

r₂ = distance of the charge from the location x = 2 , = 2 + a

hence

E₂ = k Q /r²₂

295 x 10³ = (9 x 10⁹) Q /(2 + a)²                                              eq-1

E₄ = electric field by the charge at x = 4 , = 188.8 x 10³ N/C

r₄ = distance of the charge from the location x = 4 , = 4 + a

hence

E₄ = k Q /r²₄

188.8 x 10³ = (9 x 10⁹) Q /(4 + a)²                                              eq-2

295 x 10³ /(188.8 x 10³ ) = ((9 x 10⁹) Q /(2 + a)² ) /((9 x 10⁹) Q /(4 + a)²)

295 /188.8 = (4 + a)² /(2 + a)²

a = 6 m

so location of the charge is 6 m to the left of origin

or   X = - 6 m

using eq-1

295 x 10³ = (9 x 10⁹) Q /(2 + a)²   

295 x 10³ = (9 x 10⁹) Q /(2 + 6)²   

Q = 0.0021 C

b)

E₃ = electric field by the charge at x = 3

r₃ = distance of the charge from the location x = 3 , = 3 + 6 = 9 m

hence

E₃ = k Q /r²₃

E₃ = (9 x 10⁹) (0.0021) /9²                                             

E₃ = 233.33 kN/C

so E₃ = - 233.33 kN/C

E₆ = electric field by the charge at x = 6

r₆ = distance of the charge from the location x = 6 , = 6 + 6 = 12 m

hence

E₆ = k Q /r²₆

E₆ = (9 x 10⁹) (0.0021) /12²                                             

E₆ = 131.3 kN/C

So E₆ = - 131.3 kN/C