A point particle with charge q = 4.1 C is placed on the x axis at x = 10 cm and

Question

A point particle with charge q = 4.1 C is placed on the x axis at x = 10 cm and a second particle of charge Q = 6.3 C is placed on the x axis at x = +25 cm.

(a) Determine the x and y components of the electric field due to this arrangement of charges at the point (x, y) = (10, 10) (the units here are centimeters).

Ex = __ N/C

Ey = __ N/C

(b) Determine the magnitude and direction of the electric field at this point.

magnitude:__ N/C

direction:__° counter-clockwise from the +x axis

Please type out an easy to follow step by step answer! I've posted this before and no one can solve.

My previous wrong answers:

Ex: 515.72, -210000

Ey: 378.83, 640000

Magnitude: 639.91, 2205000

Direction: 36.3, 163.13

Explanation / Answer

electric field due to a charge particle of charge q is given by

magnitude=k*q/r^2

where k=1/(4*pi*epsilon)=9*10^9

and r is distance of the point where the electric field has to be calculated from the point where the

charge is located.

direction of the electric field is given by:

if positive charge, then the direction is along the unit vector from the charged particle q

to the point where the electric field has to be calculated

if negative charge, then the direction is along the unit vector from the point where the electric field

has to be calculated towards the charged particle

Q(a):

field due to charge q:

charge is positive.

hence direction is from (-10,0) to (10,10)

direction vector=(10,10)-(-10,0)=(20,10)

distance=sqrt(20^2+10^2)=22.36 cm

unit vector=(20,10)/22.36=(0.89445, 0.4472)

magnitude of electric field=9*10^9*4.1*10^(-6)/0.2236^2=7.38*10^5 N/C

in vector notation:

electric field due to charge q=E1=7.38*10^5*(0.89445,0.4472)=(6.601*10^5,3.30*10^5) N/C

field due to charge Q:

charge is positive.

hence direction is from (25,0) to (10,10)

direction vector=(10,10)-(25,0)=(-15,10)

distance=sqrt(15^2+10^2)=18.0277 cm

unit vector=(-15,10)/18.0277=(-0.832, 0.5547)

magnitude of electric field=9*10^9*6.3*10^(-6)/0.180277^2=17.446*10^5 N/C

in vector notation:

electric field due to charge Q=E2=17.446*10^5*(-0.832,0.5547)=(-14.515*10^5,9.677*10^5) N/C

hence total field=E1+E2=(-791400,12.977*10^5) N/C

hence Ex=-791400 N/C

Ey=12.977*10^5 N/C

part b:

magnitude =sqrt(Ex^2+Ey^2)=15.1998*10^5 N/C

direction with +ve x axis=arctan(Ey/Ex)=121.376 degrees

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