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(17%) Problem 5: A charge $q = 9.932 \times 10^{-18}$ C moves with velocity $\vec{v} = v_z \hat{k} = (6670 \text{ m/s}) \hat{k}$ in a region with uniform electric and magnetic fields: $\vec{E} = E_x \hat{i} = (91.5 \text{ V/m}) \hat{i}$ $\vec{B} = B_y \hat{j} = (0.145 \text{ T}) \hat{j}$ Refer to the figure Sapp, Oliver - oliverbsapp@mail.missouri.edu ?17% Part (a) Using the symbols in the palette, not their numeric equivalents, enter an expression for the magnitude of the electric force acting on the charge. $F_E = $

          (17%) Problem 5: A charge $q = 9.932 \times 10^{-18}$ C moves with velocity
$\vec{v} = v_z \hat{k} = (6670 \text{ m/s}) \hat{k}$
in a region with uniform electric and magnetic fields:
$\vec{E} = E_x \hat{i} = (91.5 \text{ V/m}) \hat{i}$
$\vec{B} = B_y \hat{j} = (0.145 \text{ T}) \hat{j}$
Refer to the figure
Sapp, Oliver - oliverbsapp@mail.missouri.edu
?17% Part (a) Using the symbols in the palette, not their numeric equivalents, enter an expression for the magnitude of the electric force acting on the charge.
$F_E = $

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$(17%) Problem 5: A charge $q = 9.932 \times 10^{-18}$ C moves with velocity v⃗ = vz k̂ = (6670 m/s) k̂ in a region with uniform electric and magnetic fields: E⃗ = Ex î = (91.5 V/m) î B⃗ = By ĵ = (0.145 T) ĵ Refer to the figure Sapp, Oliver - oliverbsapp@mail.missouri.edu ?17% Part (a) Using the symbols in the palette, not their numeric equivalents, enter an expression for the magnitude of the electric force acting on the charge. FE =$

Added by Bethany G.

University Physics with Modern Physics

Hugh D. Young 14th Edition

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Solved by Expert Adam Moon

05/13/2022

Step 1

Step 1: To find the magnitude of the electric force acting on the charge, we need to use the formula for electric force: \[ F = q \cdot E \] where F is the force, q is the charge, and E is the electric field. Show more…

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Texts: (17%) Problem 5: A charge q = 9.932 x 10^-18 C moves with velocity v = vz k = (6670 m/s) k in a region with uniform electric and magnetic fields. Refer to the figure. Part (a) Using the symbols in the palette, not their numeric equivalents, enter an expression for the magnitude of the electric force acting on the charge. Part (b) Calculate the value in Newtons of the magnitude of the electric force acting on the charge. Part (c) What is the direction of the electric force on the charge? Answer all parts possible please and thank you :)

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Transcript

00:03 All right, so i took a look at your problem here, and i'm making it as an assumption that there is no acceleration on this.

00:12 This is two pithballs that are in a sort of equilibrium there.

00:19 They're just suspended here, repelling each other due to their electric force, and it's causing this angle to form.

00:28 So it looks like you got the force diagram correct.

00:33 I reset mine up just so i can talk about how i determined what to write here.

00:41 So if they are truly in equilibrium, then the net force is zero in this scenario for both of these pitfalls.

00:50 Now, that means that the x component adds up to zero for its net force, and the y component is also zero.

00:58 Now, what components are causing or canceling out in the x component? it's the electric force and the x component of the ft.

01:11 So if i was to look at a triangle of ft, well, here, i'll just write it on the force diagram.

01:24 I have this triangle.

01:27 And so this is ftx here.

01:30 You can erase this.

01:36 Ft in the x component.

01:37 Ft and the y component.

01:40 So ftx, which is sine theta ft, must cancel out the electric force here of fe.

01:52 Those must add up to zero.

01:55 So they must cancel each other.

01:57 They must be equal to each other in terms of magnitude.

02:01 They're just in opposite directions.

02:04 Same is true about the y component.

02:06 There we've got the cosine theta of ft minus fg equals zero.

02:12 So i don't know if that was what was, if it was that simple of a problem.

02:18 I don't know if there was more to it than that.

02:22 They were just adding up to zero...

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