A mettalic sphere of radius a carries a charge Q. It is brought in contact with an uncharged sphere of radius b . what will be the charge on the first sphere.
My work
I eliminated 2 options by substitution and realised that it won't be their average as they don't have same charge capacities
bad
they will have equal charge densities when touched, so current wont flow.
Charge densitya=q1/4PIa^2
charge densityb=q2/4PIb^2
remember q1+q2=Q You are looking for q1.
set them equal, and you get
q1/a^2=q2/b^2
q1=(a/b)^2 (Q-q1) let a/b=k
q1=kQ-kq1
q1=kQ/(1+k)
good
Well, let's see here. Two spheres, one charged and one uncharged, are brought into contact. The charge on the first sphere, let's call it the "electric" sphere, is denoted as Q. The radius of this sphere is given as a.
Now, when the electric sphere is brought into contact with the uncharged sphere, there is a transfer of charge between them. This happens because charges try to balance out. So, some charge from the electric sphere is transferred to the uncharged sphere.
However, the amount of charge transferred depends on the charge capacities of the two spheres, which are determined by their sizes (radius). The larger the sphere, the greater its charge capacity.
So, if the radius of the uncharged sphere is denoted as b, and b is greater than a, then the charge transferred from the electric sphere to the uncharged sphere will depend on their respective charge capacities.
In other words, I'm sorry to say, there's no quick and simple way to determine the exact charge on the electric sphere after contact, just based on the given information. It would require a more detailed calculation considering the charge capacities and the transfer of charge.
But hey, at least we've eliminated a couple of options based on substitution and reasoning, right? That's progress!
To find the charge on the first sphere after it is brought in contact with the uncharged sphere, we can use the concept of charge conservation.
When the two spheres are brought in contact, some charge will transfer between them until they reach the same potential. This means that the total charge before and after the contact remains the same.
The charge on the first sphere before contact is Q. The charge on the uncharged sphere is initially 0.
After contact, the charge distributes itself between the two spheres based on their size and charge capacities. The charge capacity of a conductor is proportional to its surface area.
The charge on the first sphere (with radius a) will redistribute itself onto the combined sphere (with radius a+b) according to their relative areas.
To find the charge on the first sphere, you can use the formula:
Q1/Q2 = (A1/A2)
Where Q1 is the charge on the first sphere, Q2 is the charge on the combined sphere after contact, A1 is the surface area of the first sphere, and A2 is the surface area of the combined sphere.
The surface area of a sphere is given by the formula:
A = 4πr^2
So, the ratio of the surface areas can be expressed as:
A1/A2 = (4πa^2) / (4π(a+b)^2)
Simplifying this expression, we get:
A1/A2 = (a^2) / (a+b)^2
Now, since the total charge is conserved, we have:
Q = Q1 + Q2
Rearranging this equation, we get:
Q1 = Q - Q2
Using the expression for A1/A2, we can write:
Q1 = Q - (Q * (a^2) / (a+b)^2)
Simplifying further, we get:
Q1 = Q * (1 - (a^2) / (a+b)^2)
Therefore, the charge on the first sphere after contact is Q * (1 - (a^2) / (a+b)^2).