Question

8) A negative charge, if free, tries to move A) in the direction of the electric field. B) from low potential to high potential. C) from high potential to low potential. D) away from infinity. E) toward infinity. 9) Gaussian surfaces A and B enclose the same positive charge +Q. The area of Gaussian surface A is three times larger than that of Gaussian surface B. The flux of electric field through Gaussian surface A is A) nine times larger than the flux of electric field through Gaussian surface B B) unrelated to the flux of electric field through Gaussian surface B. C) equal to the flux of electric field through Gaussian surface B. D) three times smaller than the flux of electric field through Gaussian surface B 10) If the electric field is in the positive x direction and has a magnitude given by E- C, where Cis a constant, then the electric potential is given by V

Answer 1

Ans. 8 : A negative charge when free moves from lower potential to higher potential.

Because negative charge has higher potential energy on lower potential and lower potential energy on higher potential.

   i.e P = qV (q is charge and V is potential whereas P is potential energy)

let we have a negative charge -1C and potential 1V and 2V

   at 1V potential energy = -1J and at 2V potential energy = -2J

So the charge will move from higher potential energy to lower potential energy.

Ans. 9 : Option C is correct. The flux through Surafce A will be equal to flux through surface B.

Because we know that Electric flux =
A negative charge when free moves from lower potential to higher potential. Because negative charge has higher potential energy on lower potential and lower potential energy on higher potential. i.e P = q V (q is charge and V is potential whereas P is potential energy) let we have a negative charge - 1 C and potential 1 V and 2 V at 1 V potential energy = -1 J and at 2 V potential energy = -2 J So the charge will move from higher potential energy to lower potential energy. Option C is correct. The flux through Surafce A will be equal to flux through surface B. Because we know that Electric flux = emptyset = Q_inclosed\epsilon^0 Electric flux only depends on charge inclosed not on area. Electric field does depend on gaussian area but not the flux. We know the relation between electric field and electric potential is E = -dV/dx So, dV = -E dx Putting value of E = C x2 and integrating both sides
_inclosed

Electric flux only depends on charge inclosed not on area.

Electric field does depend on gaussian area but not the flux.

Ans. 10 : We know the relation between electric field and electric potential is

  

So,    dV = -Edx

   Putting value of E = Cx² and integrating both sides

  

   So after integrating we get

V = - Cx³/3

So option D is correct.

Hope it helps....

Plz leave a thumbs up