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Nitella [24]
3 years ago
13

A particle leaves the origin with a speed of 3.6 times 106 m/s at 34 degrees to the positive x axis. It moves in a uniform elect

ric field directed along positive y axis. Find Ey such that the particle will cross the x axis at x

Physics
1 answer:
gogolik [260]3 years ago
3 0

Answer:

1.2X10^3NC

Explanation:

Pls see attached file

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Electric fields are vector quantities whose magnitudes are measured in units of volts/meter (V/m). Find the resultant electric f
Vikentia [17]

Answer:

The resultant field will have a magnitude of 241.71 V/m, 30.28° to the left of E1.

Explanation:

To find the resultant electric fields, you simply need to add the vectors representing both electric field E1 and electric field E2. You can do this by using the component method, where you add the x-component and y-component of each vector:

E1 = 99 V/m, 0° from the y-axis

E1x = 0 V/m  

E1y = 99 V/m, up

E2 = 164 V/m, 48° from y-axis

E2x = 164*sin(48°) V/m, to the left

E2y = 164*cos(48°) V/m, up

Ex: E_{1_{x}} + E_{2_{x}} = 0 V/m - 164 *sin(48) V/m= -121.875 V/m\\Ey: E_{1_{y}} + E_{2_{y}} = 99 V/m + 164 *cos(48) V/m = 208.74 V/m\\

To find the magnitude of the resultant vector, we use the pythagorean theorem. To find the direction, we use trigonometry.

E_r = \sqrt{E_x^2 + E_y^2}= \sqrt{(-121.875V/m)^2 + (208.74V/m)^2} = 241.71 V/m

The direction from the y-axis will be:

\beta = arctan(\frac{-121.875 V/m}{208.74 V/m}) = 30.28° to the left of E1.

6 0
4 years ago
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