Question

A point charge Q1=−2 μC is located at x=0, and a point charge Q2=+8 μC is placed at x=−0.5 m on the x-axis of a cartesian coordinate system.The goal of this problem is to determine the electric field, E⃗ (x)=E(x)xˆ, at various points along the x-axis.

(a)What is E(x) (in N/C) for x=-13.0 m ?

@UnkleRhaukus

Answer 1

the attchment have my wrkng about it.. i also leson to problem solvng season E have to resolve into x and y component but here just y wil not occure .. i dnt know whts prblm :(

Answer 2

you have \[\begin{align*}\vec {\mathbf E}(x)&=\vec {\mathbf E}_{Q_1}+\vec {\mathbf E}_{Q_1}\\&=K\frac{Q_1}{r_1^2}\hat{\mathbf x}+K\frac{Q_2}{r_2^2}\hat{\mathbf x}\end{align*}\] the second line should be should be \[\begin{align*}=K\frac{Q_1}{r_1^2}\hat{\mathbf r}_1+K\frac{Q_2}{r_2^2}\hat{\mathbf r}_2\end{align*}\]

Answer 3

also when you put the values in, include the sign of the charge

Answer 4

also \[r_2=-13-(-0.5)=-13+0.5\]

Answer 5

Q1=-2*10^-6 so when in put Q1 so i must put "-2*10^-6" ??
but here we are taking magnitude then why i must put neg sign ??

Answer 6

, take the magnitude at the end ,

Answer 7

i put r1=(-13.0)^2
r2=-13+0.5=-12.5
r2=(-12.5)^2
Q1=2*10^-6
Q2=8*10^-6
k=9*10^9

it gve me ans of 567.306

which is wrng :(

Answer 8

\[Q_1=-2[\text{µC}]=-2\times10^{-6} [\text C]\]

Answer 9

when i put Q1=-2*10^-6 it gve me 354.294 ans which is wrng :(

Answer 10

can i see your updated working ,

Answer 11

okay let me atch file

Answer 12

the force should be negative

Answer 13

cant understnd :(

Answer 14

\[\hat{\mathbf r}_1=\hat{\mathbf x}\]\[\hat{\mathbf r}_2=-\hat{\mathbf x}\]

Answer 15

:(

Answer 16

-354.3

Answer 17

corect