+91-8076753736 support@wiredfaculty.com
logo
logo

Electric Potential

Sponsor Area

Question
ICSEENIPH12029769
Wired Faculty App

Electrostatic potential energy stored in capacitor C3. (c) ‘n’ cells, each of emf ’e’ and internal resistance ‘r’ are joined in series to form a row. ‘m’ such rows are connected in parallel to form a battery of N = mn cells. This battery is connected to an external resistance ‘R’

(i)    What is the emf of this battery and how much is its internal resistance?

(ii)    Show that current ‘I’ flowing through the external resistance ‘R’ is given by :

straight I space equals space fraction numerator Ne over denominator mR space plus space nr end fraction

Solution

Emf = ne
 
Equivalent Internal resistance = R'
straight R apostrophe space equals space nr over straight m ii right parenthesis thin space Current space apostrophe straight I apostrophe space flowing space through space the space external space space space space space resistance space straight R space is space given space by comma space straight I space equals space fraction numerator EMF over denominator Total space resistance end fraction space equals space fraction numerator ne over denominator straight R space plus space begin display style nr over straight m end style end fraction straight I space equals space fraction numerator straight n space em over denominator mR space plus space nr end fraction space space equals space fraction numerator Ne over denominator mR space plus space nr end fraction space left parenthesis because space nm space equals space straight N right parenthesis

Sponsor Area

Question
ICSEENIPH12029814
Wired Faculty App

Three point charges Q1 = 25μ C, Q2 = 50μ C, and Q3 = 100μ C are kept at the comers A, B and C respectively of an equilateral triangle ABC having each side equal to 7.5 m. Calculate the total electrostatic potential energy of the system.

Solution

Electrostatic PE of the system: 
                        

straight E subscript straight P. straight E end subscript space equals space fraction numerator 1 over denominator 4 πε subscript straight o end fraction open square brackets fraction numerator 25 space straight x 10 to the power of negative 6 end exponent straight x space 50 straight x 10 to the power of negative 6 end exponent over denominator 7.5 end fraction space plus fraction numerator 25 space straight x 10 to the power of negative 6 end exponent straight x space 100 space straight x 10 to the power of negative 6 end exponent over denominator 7.5 end fraction plus fraction numerator 25 space straight x 10 to the power of negative 6 end exponent straight x space 100 space straight x 10 to the power of negative 6 end exponent over denominator 7.5 end fraction space close square brackets straight P. straight E space of space the space system space is comma space straight P. straight E space equals space 9 space straight x 10 to the power of 9 space open square brackets fraction numerator 25 space straight x space 50 straight x 10 to the power of negative 12 end exponent over denominator 7.5 end fraction space plus fraction numerator 25 space straight x space 100 space straight x 10 to the power of negative 12 end exponent over denominator 7.5 end fraction plus fraction numerator 25 space straight x space 100 space straight x 10 to the power of negative 12 end exponent over denominator 7.5 end fraction space close square brackets space space space space space space equals space fraction numerator 9 space straight x 10 to the power of 9 space straight x 10 to the power of negative 12 end exponent over denominator 7.5 end fraction space open square brackets 1250 space plus space 2500 space plus space 5000 close square brackets space space space space space space equals fraction numerator 9 space straight x 10 to the power of negative 3 end exponent over denominator 7.5 end fraction space open square brackets 8750 close square brackets space space space space space space equals space 10.5 space Joules

Question
ICSEENIPH12029815
Wired Faculty App

Obtain and expression for equivalent capacitance C, when three capacitors having capacitance C1 , C2 and C3 are connected in series.

Solution



Due to the applied p.d V,charge supplied by the battery is Q and due to induction,charge developed on each battery is Q applied p d are v1 ,v2 and v3 . 

V = V1 + V2 + V3                              ...(i)

Q = i1V1 = i2V2 = i3V3                    ---(ii)
In place of the above arrangement a single capacitor C equ is imagined which can withstand the applied P.D v and can store the same magnitude of charge Q.

Q =CequV                                ---(iii)

From Eq. (i) + Eq. (iii) 

Q = Cequ (V1+V2+V3)               ... (iv)

From Eq. (ii) + Eq. (iv)
straight Q over straight C subscript eq space equals space Q over C subscript 1 space plus space Q over C subscript 2 space plus space Q over C subscript 3 Q over C subscript e q u i v. end subscript space equals space 1 over C subscript 1 space plus space 1 over C subscript 2 space plus space 1 over C subscript 3

Question
ICSEENIPH12029893
Wired Faculty App

In an electric dipole, what is the locus of a point of zero potential?

Solution

The locus of the point of zero potential of an electric dipole is straight line drawn at the middle of the electric dipole.

Sponsor Area

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation