Electric Charges and Fields

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Question 1

A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. The electric potential at the point O lying at a distance L from the end A is

  • fraction numerator straight Q over denominator 8 πε subscript straight o straight L end fraction
  • fraction numerator 3 space straight Q over denominator 4 space πε subscript straight o straight L end fraction
  • fraction numerator straight Q over denominator 4 πε subscript straight o straight L space In space 2 end fraction
  • fraction numerator straight Q space In space 2 over denominator 4 πε subscript straight o straight L end fraction

Solution

D.

fraction numerator straight Q space In space 2 over denominator 4 πε subscript straight o straight L end fraction
dv space equals space kdQ over straight x
straight V space equals space integral subscript straight L superscript 2 straight L end superscript dV
straight V space equals space integral subscript straight L superscript 2 straight L end superscript kdQ over straight x space equals integral subscript straight L superscript 2 straight L end superscript fraction numerator straight k space open parentheses begin display style straight Q over straight L end style close parentheses space dx over denominator straight x end fraction
space equals space fraction numerator straight Q over denominator 4 πε subscript straight o straight L end fraction integral subscript straight L superscript 2 straight L end superscript open parentheses 1 over straight x close parentheses space dx space equals space fraction numerator straight Q over denominator 4 πε subscript straight o straight L end fraction left square bracket log subscript straight e space equals space straight x right square bracket subscript straight L superscript 2 straight L end superscript
space equals space fraction numerator straight Q over denominator 4 πε subscript straight o straight L end fraction left square bracket log subscript straight e 2 straight L space minus space log subscript straight e straight L right square bracket
space equals space fraction numerator straight Q over denominator 4 πε subscript straight o straight L space end fraction open square brackets log subscript straight e space fraction numerator 2 straight L over denominator straight L end fraction close square brackets space equals space fraction numerator straight Q over denominator 4 πε subscript straight o straight L end fraction space In space left parenthesis 2 right parenthesis
Question 2

A charged ball B hangs from a silk thread S which makes an angle θ with a large charged conducting sheet P, as shown in the figure. The surface charge density σ of the sheet is proportional to

  • cos θ

  • cot θ

  • sin θ

  • tan θ

Solution

D.

tan θ

tan space fraction numerator qσ over denominator left parenthesis 2 straight epsilon subscript straight o right parenthesis mg end fraction
Question 3

A charged particle q is shot towards another charged particle Q which is fixed, with a speed v it approaches Q upto a closest distance r and then returns. If q were given a speed 2v, the closest distances of approach would be

  • r

  • 2r

  • r/2

  • r/4

Solution

D.

r/4

By principle of conservation of energy
1 half mv squared space equals space kqQ over straight r space..... space left parenthesis straight i right parenthesis

Finally comma space 1 half space straight m left parenthesis 2 straight v right parenthesis squared space equals space fraction numerator begin display style KqQ end style over denominator straight r squared end fraction space... left parenthesis ii right parenthesis
Equation space left parenthesis straight i right parenthesis divided by left parenthesis ii right parenthesis
1 fourth space equals space fraction numerator straight r apostrophe over denominator straight r end fraction
rightwards double arrow space straight r apostrophe space equals space fraction numerator begin display style straight r end style over denominator 4 end fraction

Question 4

A charged particle with charge q enters a region of constant, uniform and mutually orthogonal fields straight E with rightwards arrow on top space and space straight B with rightwards arrow on top, with a velocity straight v with rightwards arrow on top perpendicular to both straight E with rightwards arrow on top space and space straight B with rightwards arrow on top , and comes out without any change in magnitude or direction of straight v with rightwards arrow on top .Then

  • straight v with rightwards arrow on top space equals space straight E with rightwards arrow on top space straight x space straight B with rightwards arrow on top space divided by straight B squared
  • straight v with rightwards arrow on top space equals space straight B with rightwards arrow on top space straight x space straight E with rightwards arrow on top divided by straight B squared
  • straight v with rightwards arrow on top space equals space straight E with rightwards arrow on top space straight x space straight B with rightwards arrow on top divided by straight E squared
  • straight v with rightwards arrow on top space equals space straight B with rightwards arrow on top space straight x space straight E with rightwards arrow on top divided by straight E squared

Solution

A.

straight v with rightwards arrow on top space equals space straight E with rightwards arrow on top space straight x space straight B with rightwards arrow on top space divided by straight B squared
Question 5

A long cylindrical shell carries positive surface charge  in the upper half and negative surface charge  in the lower half. The electric field lines around the cylinder will look like figure given in: (figures are schematic and not drawn to scale)

Solution

D.

Field lines should originate from a positive charge and terminate negative charge.