What is the magnitude of the electric field at point P, located at (6.50 cm, 0),

Question

What is the magnitude of the electric field at point P, located at (6.50 cm, 0), due to Q1 alone?
6.02×106 N/C

What is the x-component of the total electric field at P?
1.25×107 N/C

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 = 3.90 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?

wo charges, Q1= 3.90 C, and Q2= 5.60 C are located at points (0,-4.00 cm ) and (0,+4.00 cm), as shown in the figure.

Explanation / Answer

Y - component of total electric field at P is

E will make an angle of tan^-1[4/6.50] = 31.6 degree with positive x-axis

E = EQ1+EQ2

E = (1/4*pi*epsilono )* ( Q1/r1^2-Q2/r2^2) *Sin31.6

r1 =r2 = sqrt(6.5^2+4^2) = 7.632 cm = 0.07632 m

E = 9*10^9 * (1/0.005825)* (1.7*10^-6) = 1.376*10^6 N/C

TOTAL FIELD AT P IS

E = Sqrt( 6.02^2+1.376^2) *10^6

E = 6.175 *10^6 N/C

If Both Charges are Equal...

Y- Component of electric Field is Zero

X- Components will add up, EX = E = (1/4*pi*epsilono )* 2*Q *Cos 31.6 / 0.005825

EX = 10.264*10^6 N/C

Total Field is E = 10.26*10^6 N/C

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