Calculating Forces on Charged Particles Using Vector Components

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To calculate the force on a +8 micro-Coulomb charge, the formula Fe = q*E is used, resulting in a force of 1.6 x 10^-4 N. The electric field strength provided is 20 N/C, and the force needs to be expressed in i, j, k format. The calculation involves using trigonometric functions: 1.6 x 10^-4 * cos(60) for the i component and 1.6 x 10^-4 * sin(60) for the j component. This method simplifies vector addition by separating horizontal and vertical components, allowing for independent analysis of motion in perpendicular directions. Understanding these components is crucial for solving problems involving forces on charged particles.
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1. Find in the i, j, k format, the force on the +8 micro-Coulomb charge (pic att'd)

F = 1.6 x 10^-4 N
no clue where to start

soa9tf.jpg


thx
 

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Use the formula for electric field causing a force on a charge: Fe = q*E
 
got that. that's where the 1.6 x 10^-4 came from.
 
Are you finished then, or do you need to add the electric field due to the other charge? A good idea to show formula and numbers - the answer is not much of a clue as to what you have done.
 
electric field strength is on the image - 20 N/C

the issue here is, they want it in i, j k format
 
Okay.
1.6 x 10^-4 * cos(60) i + 1.6 x 10^-4 * sin(60) j + 0k
 
thanks delphi. can you explain the cos and sin portion a bit? i never really could understand why those are used to express a single side of a triangle
 
triangle.jpg

Expressing a vector as a horizontal part plus a vertical part makes it very convenient to add vectors (just add the h and v parts separately without worrying about angles). Also, in something like a trajectory problem the motion in two perpendicular directions is independent, so you can do the horizontal part of the problem without worrying about the vertical part.
 
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