Finding the magnitude and direction in a electric field.

[Trap_Shooter]

Three point charges are placed at the vertices of an equilateral triangle (of side .2m). The triangle is setup on an X-Y plain. On the top of the triangle on the Y-axis the charge is q7= -7μC, the positive X-axis the charge is q3= 3μC, and on the -X-axis the charge is q2= 2μC.

Part A).
What is the magnitude and the direction of the electric field at the origin?

Part B).
What would be the acceleration (magnitude and direction) of an electron placed at the origin?

Part C).
Would the acceleration (magnitude and direction) of the proton be different than that of the electron? Explain.

 

[haruspex]

As per forum rules, you must post any equations you have that you believe may be relevant and your own attempt at solution, as far as you got.

 

[13characters]

I'm going to parasite on this thread, because it deals with a similar topic.

Problem 2.7 Find the electric field a distance z from the center of a spherical surface of radius R, which carries a uniform charge density δ. Treat the case z < R as well as z > R. Express your answer in terms of the total charge q on the sphere.

the way i approached it was use the electric field equation,

E = [itex]\frac{1}{4\piε}[/itex] [itex]\int[/itex] [itex]\frac{1}{|r|^2}[/itex] [itex]\widehat{r}[/itex]dq

for which i substituted [itex]\widehat{r} = cosψ = \frac{z - Rcos\theta}{r}[/itex] and for [itex] |r| = √R^2sin^2θ + (z - Rcosθ)^2[/itex] and [itex] dq = δda = δr^2sinθdθd\phi[/itex]

where R is the radius of the sphere upon whos surface the charge is distributed.

Now i substitute all of these wonderful thing into the electric field equation, all giddy to finally solve it :D Integrating over the entire surface of the sphere, that is from 0 to [itex]\pi [/itex] for [itex]d\theta[/itex] and 0 to [itex]2\pi [/itex] for [itex]d\phi[/itex]

E = [itex]\frac{1}{4\piε} \int \frac{1}{ (√R^2sin^2θ + (z - Rcosθ)^2)^2 }\frac{z - Rcos\theta}{(√R^2sin^2θ + (z - Rcosθ)^2)}δr^2sinθdθd\phi[/itex]

[itex]\int d\phi = 2\pi[/itex] nothing exciting there, parametrize [itex] u = cosθ, du = -sinθ dθ [/itex]

E = [itex]\frac{1}{4\piε} \int \frac{δ(2\pi r^2sinθ)(z - Rcos\theta)}{ (√R^2sin^2θ + (z - Rcosθ)^2)^3 }dθ \Rightarrow E = \frac{1}{4\piε} \int \frac{δ(2\pi r^2sinθ)(z - Rcos\theta)}{ (√R^2 + z^2 - 2Rzcosθ)^3 }dθ \Rightarrow E = -\frac{1}{4\piε} \int \frac{δ(2\pi r^2)(z - Ru)}{ (√R^2 + z^2 - 2Rzu)^3 }du[/itex]

at this point i pretty much just hur dur, try to integrate by partial fractions, and get nowhere because its like no other I've met before.

I realize this is a pretty standard integral, being an inverse cosine law and all. Any online resources?

Would someone please hint as to how to solve this, have i made any mistakes.

 

[13characters]

Three point charges are placed at the vertices of an equilateral triangle (of side .2m). The triangle is setup on an X-Y plain. On the top of the triangle on the Y-axis the charge is q7= -7μC, the positive X-axis the charge is q3= 3μC, and on the -X-axis the charge is q2= 2μC.

step one, draw the question. apply principle of superposition; consider one charge at a time, then sum individual forces.

Part A
think about which force vectors are going to cancel at the origin.

Part B
use the result from part A :D

Part C

is pretty vanilla. opposites attract.

 

[Nickie]

Part A). To find the magnitude and direction of the electric field at the origin, we can use the formula for electric field: E = kq/r^2, where k is the Coulomb's constant, q is the charge, and r is the distance from the charge to the origin.

First, we need to find the distance from each point charge to the origin. Since the triangle is equilateral, all sides are equal and the distance from each charge to the origin is 0.2m.

Using the values given, we can calculate the electric field from each charge as follows:
- q7: E = (9x10^9 Nm^2/C^2)(-7x10^-6 C)/(0.2m)^2 = -220.5 N/C
- q3: E = (9x10^9 Nm^2/C^2)(3x10^-6 C)/(0.2m)^2 = 450 N/C
- q2: E = (9x10^9 Nm^2/C^2)(2x10^-6 C)/(0.2m)^2 = 300 N/C

To find the total electric field at the origin, we can use vector addition. Since the electric field from q7 is negative and the electric fields from q3 and q2 are positive, the total electric field will have a direction towards the positive X-axis.

The magnitude of the total electric field can be found by taking the square root of the sum of the squares of the individual electric fields:
E = √(220.5^2 + 450^2 + 300^2) = 570.4 N/C

Therefore, the magnitude of the electric field at the origin is 570.4 N/C and the direction is towards the positive X-axis.

Part B). To find the acceleration of an electron placed at the origin, we can use the formula for electric force: F = Eq, where E is the electric field and q is the charge of the particle.

Since the charge of an electron is -1.6x10^-19 C, the electric force on the electron can be calculated as follows:
F = (570.4 N/C)(-1.6x10^-19 C) = -9.1x10^-17 N

Using Newton's second law (F = ma), we can find