I have been given the answers, but I need to show my work on how I arrive at the

Question

I have been given the answers, but I need to show my work on how I arrive at the answers. Thank you for any help you can give me

I have been given the answers, but I need to show my work on how I arrive at the answers. Thank you for any help you can give me An electron in a certain region of space experiences a force of 4.8 times 10-15 N due North from the electric field there. Find the magnitude and direction of the electric field. What is the magnitude and direction of the electric force on an ion Mg++ placed at the same location as the previous problem? What is the acceleration of the ion in the previous problem, given that its mass number is 24? A point charge of 20nC is placed on the x-axis at the point x = 2. Find the x-component of the electric field Ex at x = 4 and x = -1 where the coordinates are in meters. Sketch Ex as a function of x. Repeat the previous problem if the point charge is - 20nC. The next three problems refer to two point charges A and B on the x-v plane. With coordinates measured in meters, A is at (-1,0) and carries charge - 5nC. B is at (2,0) and carries + 4nC. Find the x-component of the electric field Ex at the points (0,0), (-4,0) and (1,0) and plot Ex as a function of x. Find the x-coordinate of the point on the x-axis where the electric field is zero. Find the magnitude of the electric field at the point (2.3). On the x-y plane, two equal point charges q are located at the points (a.0) and (-a,0). Find the magnitude of the electric field at the point (0,y), Find the value of y where the electric field is maximum. Sketch the electric field as a function of y. An electric dipole consists of a point charge -q at the point (0,a) and a point charge q at the point (0,-a). Find the elecnic field at the point (x,0). Three point charges -q. -q. and 2q occupy three corners of a square of side a as shown. Find the magnitude and direction of the electric field at (a) the unoccupied comer, and (b) the center. Three point charges -q. -q. and 2q occupy the vertices of an equilateral triangle of side a. Find the electric field at the center of the triangle. A water droplet of radius 0.01mm remains stationary in air. If the downward electric field is 150N/C, how many excess elections must the droplet have? A calcium ion Ca++ enters with zero velocity a region of uniform electric field of 200N/C pointing east. What is its speed after 50ms? (Th symbol of calcium atom is 20Ca40) Ail electron travelling east with velocity 7.0 times 105 m/s enters the same electric field region as in the previous problem. How far does it go before being turned backward? Repeat the previous problem if the initial velocity of the electron is at 30' from the east. An alpha particle travelling horizontally at 50km/s enters a region where a downward uniform electric field of300N/C is present. After travelling forward 15cm. how far has it moved downward? Answers: 3.0times104N/C, south 9.6times10-15N 2.4times1011m/s2 45N/C, - 20N/c -45N/C, 20N/c -54N/C, 4.0N/c, -47.3N/C 27.4m 2.85N/C 2kqy/(a2 + y2)3/2, 2kqa/(a2 + x2)3/2 ( -1)kq/a2 directed toward 2q, 4kq/a2 9kq/a2 107 4.8times107m/s 7.0mm 5.3mm 6.5cm

Explanation / Answer

You are only allowed to ask one question per post on Chegg, and I am really only allowed to answer 1 question. I will answer a few for you here, but if you want more answered, and not be in violation of policy, repost each question individually with a minimum of 300 points.

I will answer 5 questions here which should be worth 1500 points, but I will give you the benefit of the doubt, and they are sort of related.

Anyway...

Number 1)

Apply F = qE

(4.8 X 10^-15) = (1.6 X 10^-19)(E)

E = 3.00 X 10^4 N/C South

(South because the field acts in the opposite direction to the force for a negative charge)

***(It would act along the direction of force for a positive charge)

Number 2)

Apply F = qE

F = (2)(1.6 X 10^-19)(3 X 10^4)

E = 9.60 X 10-15 N

Number 3)

F = ma

9.6 X 10^-15 = (24)(1.67 X 10^-27)(a)

a = 2.4 X 10^11 m/s^2

Number 4)

Apply E = kq/r^2

At the 4 m location

E = (9 X 10^9)(20 X 10^-9)/2^2

E = 45 N/C to the right, so its positive

At the -1m location

E = (9 X 10^9)(20 X 10^-9)/3^2

E = 20 N/C to the left so its negative... That is -20 N/C

Number 5)

Same idea as 4, but just the opposite charge

All vales in the formulas are the same but directions opposite

At the 4 m location

E = (9 X 10^9)(20 X 10^-9)/2^2

E = 45 N/C to the left, so its negative... That is -45 N/C

At the -1m location

E = (9 X 10^9)(20 X 10^-9)/3^2

E = 20 N/C to the right so its positive

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