Need help on everything, A through F (17%) Problem 2: A positive charge of magni

Question

Need help on everything, A through F

(17%) Problem 2: A positive charge of magnitude Q1-8.5 nC is located at the origin A negative charge Q2-9.5 nC is located on the positive x-axis at x-3.5 cm from the origin. The point P is located y 17.5 cm above charge 22 Otheexpertta.con 17% Part (a) Calculate the x-component of the electric field at point P due to charge Q1. Write your answer in units of N/C. Grade Summary Deductions Potential 0% 100% cosO cotan0 asin0 acos0 atan0 acotan0 sinhO sin() tan() ( Submissions Attempts remaining % per attempt) detailed view cosh0tanh0 c cotanh0 0 Degrees O Radians Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining: 4 Feedback: 0% deduction per feedback. 17% Part (b) Calculate the y-component of the electric field at point P due to charge Q1. Write your answer in units of N/C 17% Part (c) Calculate the y-component of the electric field at point P due to the Charge Q2. Write your answer in units of NC. 17% Part (d) Calculate the y-component of the electric field at point P due to both charges. Write your answer in units of N/C. 17% Part (e) Calculate the magnitude of the electric field at point P due to both charges. Write your answer in units of N/C. 17% Part (f) Calculate the angle in degrees of the electric field at point P relative to the positive x-axis

Explanation / Answer

We know that electric field at a point is given by:

E = k*Q/r^2, where r = distance between charge and point

Direction of electric field at point is towards the negative charge and away from positive charge

Part A.

Since Q1 is positive, So electric field will be in north-east direction away from charge Q1

E1 = k*Q1/R^2

R = distance between P and Q1 = sqrt (3.5^2 + 17.5^2) = 17.85 cm

R = 0.1785 m

Q1 = 8.5 nC

E1 = 9*10^9*8.5*10^-9/(0.1785^2)

E1 = 2400.96 N/C

Now electric field E1 in x-direction will be

E1x = E1*cos A = E1*(x/R)

E1x = 2400.96*(3.5/17.85)

E1x = 470.78 N/C

Part B.

E1y = E1*sin A = E1*(y/R)

E1y = 2400.96*(17.5/17.85)

E1y = 2353.88 N/C

Part C.

Since Q2 is negative, so electric field will be towards P in south direction (means negative y-axis)

E2 = E2y = -k*Q2/y^2

y = 17.5 cm = 0.175 m

E2y = -9*10^9*9.5*10^-9/(0.175)^2

E2y = -2791.84 N/C

Part D.

Net electric field in y-direction will be

Enet_y = E1y + E2y

Enet_y = 2353.88 - 2791.84

Enet_y = -437.96 N/C

Part E.

Enet = Enet_x + Enet_y

Enet = 470.78 i - 437.96 j

|Enet| = sqrt (470.78^2 + (-437.96)^2)

|Enet| = 643.0 N/C

Part F.

Direction = arctan (Enet_y/Enet_x)

= arctan (-437.96/470.78)

= -42.93 deg

= 42.93 deg below positive x-axis

Please Upvote. Comment below if you have doubt in any step.

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.