Question

Two charges, Q1= 3.70 pC, and Q2= 6.60 pC are located at points (0,-3.50 cm ) and (0,+3.50 cm), as shown in the figure What is the magnitude of the electric field at point P, located at (6.00 cm, 0), due to Q1 alone? 6.89×106 N/C You are correct. Previous Tries What is the x-component of the total electric field at P? 1.66x107 N/C You are correct. Previous Tries What is the y-component of the total electric field at P? 308.9 NIC Incorrect. Tries 3/20 Previous Tries Suomit Answer What is the magnitude of the total electric field at P? 1.68x107 N/C You are correct. Previous Tries Now let Q2-Q1 = 3.70 μC. 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? 1576.2 N/C Can you say anything about the y-component of the total electric field before you start calculating? submit Anewer Incorrect. Tries 1/20 Previous Tries Given the symmetric situation of the previous problem, what is the magnitude of the force on an electron placed at point P? 2.52 10-16 N Recall what the electric force will be on a charge q in the presence of an electric field E Submit Answer Incorrect. Tries 1/20 Previous Tries

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Answer #1

Given is:-

Qi = 3.70× 10-6C.Q) = 6.60× 10-6C

Now,

part-1

The distance between P and Q1 can be found out by using pythagorous theorem

thus

r = V/(0.0350)2 (0.06)2

r = 0.0695m

The electric field at P due to Q1 is

KQ1 Ei

by plugging all the values we get

9 x 10 (3.70 x 10 E1 (0.0695)2

El 6.89 100V/C

Part-2

The electric field due to charge Q2 is

9 × 109(6.60 x 10-6) (0.0695)2 E, =

which gives us

E_2 = 1.23 imes 10^7 N/C

Now,

the x component of the net electric field will be

E_x = E_1_x+E_2_x

by plugging all the values we get

E, = 6.89 × 106 × cosố + 12.3 × 10° × cosố

the value of cosine is  0.06 cos 0,0695

by plugging this value in above expression we get

0.06 0.06 Er = 6.89 × 106 × 12.3 × 106 × 0,0695 0,0695

we get

E, 1.66 × 10·N/C

Part-3

The y component of the net electric field will be

E_y = E_1_y + E_2_y

E, 6.89 × 10° × sinθ-12.3 × 10° × sinθ

the value of the sin in the triangle is given by  

sing-0.0350 0.0695

by plugging this values we get

E, 6.89 × 106 × 0.0350 0,0695- 12.3 × 106 0.0350 0.0695 12.3 × 106 ×

or

E,-- 2.7245 × 100 N/C

Part-4

The magnitude of the total electric field will be

E = sqrt{(E_x)^2+(E_y)^2}

E = V/( 1.66 × 107)2+ (-2.7245 × 100)2

which gives us

E-1.68 × 10 y/c

Part-5

If

Q_1 = Q_2 = 3.70 imes 10^{-6}C

Then

-0.00 0.0695 Etot = 2×6.89 × 100 ×

which gives us

Etot 1.19 × 10N/C

Part-6

The y components of the both E1 and E2 will cancel each other out because they are same in magnitude but opposite in the direction.

Part-7

The electric force will be

tot

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