What is the magnitude of the electric field at point P, located at (4.50 cm, 0),

Question

What is the magnitude of the electric field at point P, located at (4.50 cm, 0), due to Q1 alone?

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

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

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

Now let Q2 = Q1 = 3.20 ?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?

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

Two charges, Q1= 3.20 ?C, and Q2= 6.20 ?C are located at points (0,-4.00 cm ) and (0,+4.00 cm), as shown in the figure.

Explanation / Answer

r=sqrt[42+4.52] =6.02 cm

a)

Electric field at point P due to Q1 is

E1 =KQ1/r2 =(9*109)(3.2*10-6)/0.06022

E1=7.95*106 N/C

b)

from figure

cos(o) =4.5/6.02 =0.7475

sin(o)= 4/6.02= 0.66445

Electric field at P by Q2 is

E2 = k.Q2/r2 = (9*109).(6.2*10-6)/(0.06022)

E2 = 15.4 *106 N/C

X component is

Ex = E1 cos(0) + E2 cos (o) =(7.95*106)*0.7475 +(15.4*106)*0.8944

Ex= 17.45*106 N/C

C)

Y component is

Ey = E1 sin(o)- E2 sin(0) = (7.95*106)(0.66445) - (15.4*106)(0.66445)

Ey= -4.95*106 N/C

D)

Et = sqrt[(Ex2 + Ey2] =sqrt[(17.45*106)2+(-4.95*106)2]

Et=18.14*106N/C

E)

if Q1 =Q2 =3.2uC

then E1 = E2

and the y component of E1 and E2 cancel out .

=>Et= 2E1 cos (o)=2*(7.95*106)*(0.7475)

Et=11.89*106 N/C   

F)

Electric force

F =qEt= (1.6*10-19)(11.89*106)

F= 1.9*10-12 N

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