What is the magnitude of the electric field at point P, located at (6.00 cm, 0),

Question

What is the magnitude of the electric field at point P, located at (6.00 cm, 0), due to Q1 alone?

A   ________________

What is the x-component of the total electric field at P?

B ___________________

What is the y-component of the total electric field at P?

C _____________________

What is the magnitude of the total electric field at P?

D   _____________________

Now let Q2 = Q1 = 2.50 ?C. Note that the problem now has a symmetry that you should exploit in your solution. What is the magnitude of the total electric field at P?

E _______________________

Given the symmetric situation of the previous problem, what is the magnitude of the force on an electron placed at point P?

F_______________________

Explanation / Answer

r=sqrt[32+62] =6.71 cm

a)

Electric field at point P due to Q1 is

E1 =KQ1/r2 =(9*109)(2.5*10-6)/0.06712

E1=5*106 N/C

b)

from figure

cos(o) =6/6.71 =0.8944

sin(o)= 3/6.71= 0.4472

Electric field at P by Q2 is

E2 = k.Q2/r2 = (9*109).(5*10-6)/(0.06712)

E2 = 10 *106 N/C

X component is

Ex = E1 cos ? + E2 cos ? =(5*106)*0.8944 +(10*106)*0.8944

Ex= 13.416*106 N/C

C)

Y component is

Ey = E1 sin ? - E2 sin ? = (5*106)(0.4472) - (10*106)(0.4472)

Ey= -2.236*106 N/C

D)

Et = sqrt[(Ex2 + Ey2] = ?[(1.34*107)2+(-0.2236*107)2]

Et=13.6*106N/C

E)

if Q1 =Q2 = 2.50uC

then E1 = E2

and the y component of E1 and E2 cancel out .

=>Et= 2E1 cos (o)=2*(5*106)*(0.8944)

Et=8.944*106 N/C   

F)

Electric force

F =qEt= (1.6*10-19)(8.944*106)

F= 1.43*10-12 N

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