Question

pls answer these 2 questions.
I will mark the Answer as brainliest..

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Answer 1

answer of your question 1st


Let 3 capacitor of capacitance C1,C2,C3C1,C2,C3 be connected in parallel potential difference q is applied  across the combination . 
i.e $q=q_1+q_2+q_3q_1= C_1 Vq_2=C_2 Vq_3=C_3 Vandandq=CV\therefore CV=C_1V+C_2V+C_3VororC=C_1+C_2+C_3$
The effective capacitance is the sum of the individual capacitance.  
$\frac{1}{c} = \frac{1}{c1} + \frac{1}{c2}$

2nd answer
Electric potential can be defined in two ways,

•It is the work done by an external agent at a point to bring a unit positive charge from infinity to that point.

•It is the negative of the work done by the conservative force( which in this case is the electrostatic force) to bring a unit positive charge from infinity to that point.


VA=−W∞−>A,c.forceq0

VA=−W∞−>A,c.forceq0


We will use the second definition to derive an expression for an electric potential due to an isolated point charge.


Now suppose that you have a charge q and  q  already assembled at a point say A  

From the very definition of potential, you are bringing a charge say q0 and  q0  from infinity to a point say B   situated at a distance r   from A  

As the charge is bought from ∞  
to B, the conservative force due to charge q   opposes the work done by external agent to move q0   from ∞   to B


Let that force be labelled as Fconservative


Let the distance between the charge q   and q0  
beyond point B   be x  
as it is being bought from ∞

 
Now,


Fconservative=kqq0x2

Fconservative=kqq0x2


Let there be a small displacement dx   due to the external agent. In order to oppose the electrostatic force due to q  , some work must be done.


Let a small work dWcons   be done to displace charge q0   by dx

 
Therefore,

 dWcons=Fcons∙dx

dWcons=Fcons•dx


Since x is decreasing, apart from taking θ   from the dot product, we will take a negative sign for dx  
as well.


dW=Fcons(−dx)cos180°

dW=Fcons(−dx)cos180°  (as direction of displacement is oppsite to the conservative force)


dW=Fconsdx

dW=Fconsdx


dW=kqq0x2dx

dW=kqq0x2dx


We have to now find the work done by conservative force, for that we have to integrate both sides taking upper and lower limits as r   and ∞  
respectively.


W=∫r∞kqq0x2dx

W=∫∞rkqq0x2dx


W=kqq0∫r∞x−2dx

W=kqq0∫∞rx−2dx


W=kqq0[−1x]r∞

W=kqq0[−1x]∞r


W=−kqq0r

W=−kqq0r


Potential is the negative of the work done by the conservative force per unit charge, therefore

 VA=kqr

VA=kqr


Hence the potential at point A   is

VA=1/4πϵ0qr  

Answer 2

Answer:

Solution is attached above ...