Hot answers tagged

4

For point sources of a field or energy source, such as a charged particle, a gravitational body (which acts like a point source), or a loudspeaker on top of a tall column, the geometry of the problem controls how energy and fields distribute themselves in space. At all points that are an equal distance from a point source, the energy or field strength is ...


2

You are not completely mistaken. What you have to do to get the desired formula is a Taylor expansion of the term which contains $R/x$ and then consider the limit when $x \to \infty$. Also be careful with notation, $\sigma=Q/ (\pi R^2)$. As Andrea Di Biagio mentions in his comment, $1/(1-u) \approx 1 - u$ when $u$ is small. In your situation $u=R^2/x^2$. ...


2

Brief Summary Numerically, the mean value property of harmonic functions allows you to get an approximate solution to boundary value problems relatively quickly. Often you can improve convergence to a solution with a good initial guess, however, so analytical approaches can still be useful. Consider the limit of an infinitely long cylinder. There is a ...


2

I sort of doubt the blue jet explanation by Xeren is probable (although it is a possibility). It's just too rare and too faint. Much more likely, it's lightning far away near the horizon. Just because there are no clouds doesn't mean the light wont scatter. Sunlight makes the sky very bright blue during the day, and in the brief moment when lightning ...


1

You have added the negative sign in front of your integral and then put in the cos(180) as well. Pick one. Since your external force is opposite in direction to the force of the sphere, and you've already put the cos(180) in there, there is no need for the extra negative sign in front.


1

This looks like a homework problem, so I'll just give some ideas on how to start this problem. The electric potential of a point dipole $\vec{p} = p \hat{z}$ can be written as $$ V_\text{dip} = \frac{1}{4 \pi \epsilon_0} \frac{\vec{p} \cdot \hat{r}}{r^2} = \frac{p}{4 \pi \epsilon_0} \frac{\hat{z} \cdot \vec{r}}{r^3}. $$ The electric potential of a pair of ...


1

It's just that the sum of kinetic and potential energies are constant through all times. You can say that potential energy is zero when they are far away, so when they are approaching, the potential becomes more negative, as it is converted to kinetic. Or you can say that the potential energy is largely positive when they are far away, then reducing to zero ...


1

Here is the simple high school level answer which knzhou's answer is better than. The electric potential energy of a system of two point charges a distance $r$ apart is given by $$ E_E=\frac{kQq}{r}$$ If the charges have opposite signs than the potential energy will be a small negative value when they are far apart. As they move closer together potential ...


1

The simple 'first year physics' answer is that the potential energy goes negative. The negative potential energy cancels out the positive kinetic energy, leaving the total energy equal to zero. This might still feel unsatisfying, because it still looks like the kinetic energy is coming 'out of nowhere'. The real resolution is better. In this situation, '...


1

you are basically trying to undergo a transition from a law which is valid for static charges(or non relativistic speeds) to one which is valid for steady currents. That is why, simple differentiation is erroneous and does not include any magnetic field term in dE/dt.


1

When you calculated $dE_x$, you multiplied by $\frac{a}{\sqrt{x^2+y^2}}$. Why? Hint: try $\frac{x}{\sqrt{x^2+y^2}}$



Only top voted, non community-wiki answers of a minimum length are eligible