T-Mobile 2:28 PM » flipitphysics.com Two point charges (qs-3.2C and -7.6 uC) are

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T-Mobile 2:28 PM » flipitphysics.com Two point charges (qs-3.2C and -7.6 uC) are fixed along the x-axis, separated by a distance d 8 cm. Point P is located at what is Ex(P), the value of the x-component of the electric field produced by q1 and q2 at point P? You currently have 5 submissions for this question. Only 10 submission are You can make 5 more submissions for this question. Your submis It appears you made a power h What is EyP), the value of the y-component of the electric field produced by q1 and q2 at point P? .510 You currently have 1 submissions for this question. Only 1O submission are aflowed You can make 9 more submissions for this question Your submis: your answer has been judged correct; the exact answer is 096509.74233027 A third point charge 3-2.9 C is now positioned along the y-axis at a distance d- 8 cm from q1 as shown. What is E(P), the x-component of the field produced by al 3 charges at point P2 You currently have 4 submissions for this question. Only 10 submission are You can make 6 more submissions for this question. *Suppose all charges are now doubled (i.e., q,.-6.4 pC q2" 15.2 q3 -5.8 . how will the electric field at P change? Olts magnitude will increase by less than a factor of two and its direction will remain the same Olts magnitude will increase by less than a factor of two and its direction will change Its magnitude will double and its direction will remain the same lts magnitude will double and its direction will change You currently have 0 submissions for this question. Only TO submission are aflowed

Explanation / Answer

in the givne problem

q1 = -3.2 uC

q2 = 7.6 uC

distance d = 8 cm = 0.08 m

location of P = (d,d)

1. x component of electric field produced by Q1 and Q2 at point P = Ex

Ex = kq1*cos(45)/2d^2

Ex = -8.98*10^9*3.2*10^-6*cos(45)/2*0.08^2 = -1587454.7237637 V/m

2. the Y component of electric field at point P due to both the charges = Ey

hence

Ey = kq1*sin(45)/2d^2 + kq2/d^2

Ey = 8.98*10^9*10^(-6)(-3.2*sin(45)/2 + 7.6)/0.08^2

Ey = 9076295.2762362 V/m

3. q3 = 2.9 uC at (0,d)

now, Ex' = Ex + kq3/d^2 = -1587454.7237637 +8.98*10^9*2.9*10^-6/0.08^2 = 2481607.77 V/m

4. if all charges are doubled, the Ex and Ey component of the electric field at point P get doubled individually

hence its magnitude will double and the direction will remain the same

5. to make net field at P = 0

the charge q1 has to increase in magnitude but remain of the same sign

hec eincrease its magnitude and keep the sign same

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