In the figure, the point charges are located at the corners of an equilateral tr

Question

In the figure, the point charges are located at the corners of an equilateral triangle 27 cm on a side. Find the magnitude of the electric field in N/C at the location of q_a, given that q_b = 11.9 mu c and q_c = 5.6 mu C |E_a| = ___________ Find the direction of the electric field at q_a in degrees above the negative x-axis with origin at q_a. What is the magnitude of the force in Non q_a, given that q_a = 1.4 nC? What is the direction of the force on q_a in degrees above the negative x-axis with origin at q_a?

Explanation / Answer

let L = 27 cm = 0.27 m

a) at the location of a

Eb = k*qb/L^2 = 9*10^9*11.9*10^-6/0.27^2 = 1.47*10^6 N/c

Ec = k*qc/L^2 = 9*10^9*5.6*10^-6/0.27^2 = 0.691*10^6 N/c

angle between Eb adnEc is 120 degrees

so,

net magnetic field at point a,

Ea = sqrt(Eb^2 + Ec^2 + 2*Eb*Ec*cos(120))

= sqrt(1.47^2 + 0.691^2 + 2*1.47*0.691*cos(120))*10^6

= 1.27*10^6 N/c

b) Eax = -Eb*cos(60) - Ec*cos(60)

= -(Eb + Ec)*cos(60)

= -(1.47 + 0.691)*10^6*cos(60)

= -1.08*10^6 N/c

Eay = Eb*sin(60) - Ec*sin(60)

= (Eb - Ec)*cos(60)

= (1.47 - 0.691)*10^6*sin(60)

= 0.674*10^6 N/c

direction of Ea, theta = tan^-1(Eay/Eax)

= tan^-1(0.674/1.08)

= 32 degrees above -x axis

c) Force on qa, F = qa*Ea

= 1.4*10^-9*1.27*10^6

= 1.78*10^-3 N

d) direction : 32 degrees above -x axis

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