I want to measure the magnitude, and the direction of the electric field at point P induced by a rod that has a charge of $-22.0\mathrm{\mu C}$. The problem has been accurately dimensioned. The distance of the rod to point P has to be calculated, because the dimensions given where to the centroid of the rod. My ultimate problem is how do I know the direction of the field? Here is my work for the problem.
\begin{equation}\vec{E}=\int\frac{k\ dq}{r^2}\hat{r} \end{equation}
\begin{equation}\frac{dq}{dx}=\frac{Q}{L}=\lambda \end{equation}
\begin{align}E&=\int_{.29}^{.43} \frac{k\lambda}{x^2}dx\\ &=\frac{kQ}{L}[-\frac{1}{x}]\Bigg|_{0.29}^{.43}\\&=-1.59\times 10^6 \frac{N}{C} \end{align}
Can anyone please help explain why the answer is that the electric field is point to the right according to my physics professors answer sheet?
Electric Field and Direction of Field
definite integralselectromagnetismphysics
Best Answer
First of all, if I read the figure right, the limits of integration are $36\pm 0.14/2$, not $0.36\pm 0.14/2$. But this will only change the numerical value.In your first equation, where you define the electric field,you have two quantities that will inform you about the direction of the electric field. $\hat r$ is a vector from the charge creating the electric field toward the point where you measure the electric field. So in your case $\hat r$ points to the left. The other quantity of interest is the charge. Yours is negative, so the electric field will point in a direction opposite to $\hat r$. Therefore, in this problem the electric field points to the right.
If you want a physical intuition, the direction of the electric field is the direction of the force acting on a positively charged particle. A positive charge at $P$ is attracted by a negatively charged rod, so the direction of the electric field is to point from $P$ to the negatively charged rod