Well at dc the current won't limit. For idealized model the current should increase linearly. If at t=0, an emf \mathcal{E} is suddenly applied to a superconductor element having inductance L, then there being no resistance, the voltage won't drop across the element implying that the back emf L...
Indeed, its a matter of convention. Usually in physics we take the work done by the gaseous system as positive. The motivation behind this convention comes the definition of work. When the system expands, its exerting some force in the outward direction and the displacement of the boundary is...
It's probably much easier if you take your +x axis along the line through the origin inclined to the horizontal at the given angle \beta and the +y axis perpendicular to it, like in the figure.
Then, we have the acceleration components:
a_x=-g\sin\beta,\quad a_y=-g\cos\beta
and the initial...
Yes the answer for the problem would be 2.90. In my last post I found the distance from q2 of the point where the field is zero. The problem asks for the coordinates, i.e. the distance from the origin which is where q1 sits. So the answer should be L+s = 2.90 L
So u fill in 2.90.
For the second one u need not use derivatives. We need the max of
\dfrac{x}{(x^2+R^2)^{3/2}}
Since the expression is odd w.r.t. x, we may as well assume x>0. Then the given expression
=\dfrac{1}{\left(\dfrac{x^2}{x^{2/3}}+ \dfrac{R^2}{x^{2/3}}\right)^{3/2}}= \dfrac{1}{\left(x^{4/3}+...
As should be clear by now, the field can be zero only on the right of q2. Suppose it is zero at a distance s from it. Then, at that point the magnitude of the fields due to the two charge must be same. As such we get (after canceling \dfrac{1}{4\pi\epsilon_0} on both side)...
First, let's have a look at the effect an external field creates on an otherwise spherical atom. When the atom is placed in the external field, the positive charges tend to move along the direction of the field while negative electron gas cloud in the opposite direction. If the field is strong...