JEE Class main Answered

Solve the question in photo showing each step clearly

Asked by Vidushi412 | 10 Apr, 2019, 09:00: AM

Expert Answer

If q is positive and given small displacement x along the line joining two charges then

F subscript n e t end subscript equals fraction numerator K Q q over denominator open parentheses a plus x close parentheses squared end fraction minus fraction numerator K Q q over denominator open parentheses a minus x close parentheses squared end fraction O n space S o l v i n g space w e space g e t F subscript n e t end subscript equals fraction numerator negative 4 K Q q a x over denominator open parentheses a squared minus x squared close parentheses squared end fraction F subscript n e t end subscript equals fraction numerator negative 4 K Q q a x over denominator a to the power of 4 open parentheses 1 minus begin display style x over a squared end style squared close parentheses squared end fraction A s space x less than less than a F subscript n e t end subscript equals fraction numerator negative 4 K Q q x over denominator a cubed end fraction F subscript n e t end subscript equals negative open parentheses fraction numerator 4 K Q q over denominator a cubed end fraction close parentheses x T h e r e f o r e comma omega equals square root of k over m end root equals square root of bevelled fraction numerator fraction numerator 4 K Q q over denominator a cubed end fraction over denominator m end fraction end root 2 πf equals square root of bevelled fraction numerator fraction numerator 4 KQq over denominator straight a cubed end fraction over denominator straight m end fraction end root straight f subscript straight a equals fraction numerator 1 over denominator 2 straight pi end fraction square root of bevelled fraction numerator fraction numerator 4 KQq over denominator straight a cubed end fraction over denominator straight m end fraction end root

If q is negative and is given a small lateral displacement

T h e space f o r c e space a c t i n g space o n space minus q space w h e n space d i s p l a c e d space t h r o u g h space x space i s space g i v e n space b y F subscript n e t end subscript equals 2 F cos theta F subscript n e t end subscript equals begin inline style fraction numerator negative 2 K Q q over denominator open parentheses a squared plus x squared close parentheses end fraction end style C o s theta F subscript n e t end subscript equals negative fraction numerator 2 K Q q over denominator open parentheses a squared plus x squared close parentheses end fraction x over open parentheses a squared plus x squared close parentheses to the power of begin display style bevelled 1 half end style end exponent F subscript n e t end subscript equals negative begin inline style fraction numerator 2 K Q q x over denominator open parentheses a squared plus x squared close parentheses to the power of begin display style 3 over 2 end style end exponent end fraction begin display style blank end style end style F subscript n e t end subscript equals negative begin inline style fraction numerator 2 K Q q x over denominator a cubed open parentheses plus begin display style x over a squared end style squared close parentheses to the power of begin display style 3 over 2 end style end exponent end fraction end style

x less than less than less than a F subscript n e t end subscript equals fraction numerator negative 2 K Q q x over denominator a cubed end fraction F subscript n e t end subscript equals negative open parentheses fraction numerator 2 K Q q over denominator a cubed end fraction close parentheses x omega equals square root of k over m end root equals square root of fraction numerator 2 K Q q over denominator a cubed m end fraction end root 2 πf equals square root of fraction numerator 2 KQq over denominator straight a cubed straight m end fraction end root straight f subscript straight b equals fraction numerator 1 over denominator 2 straight pi end fraction square root of fraction numerator 2 KQq over denominator straight a cubed straight m end fraction end root

The ratio

fraction numerator square root of 2 f subscript a over denominator f subscript b end fraction

equals square root of 2 cross times fraction numerator begin display style fraction numerator 1 over denominator 2 straight pi end fraction end style square root of begin display style fraction numerator 4 K Q q over denominator a cubed m end fraction end style end root over denominator begin display style fraction numerator 1 over denominator 2 straight pi end fraction end style square root of begin display style fraction numerator 2 K Q q over denominator a cubed m end fraction end style end root end fraction equals 2

Answered by Utkarsh Lokhande | 10 Apr, 2019, 05:09: PM

JEE Class main Answered

Application Videos

Elastic potential energy_Restoring force and spring con ...

Expression of Elastic potential energy_2

Potential Energy Curve

Work done by static friction

Work done in conservative field - 1

CALL US / WHATSAPP

Please Login or Sign up to continue!

JEE Class main Answered

Application Videos

Elastic potential energy_Restoring force and spring con ...

Expression of Elastic potential energy_2

Potential Energy Curve

Work done by static friction

Work done in conservative field - 1

Choose Filters

Open In Ask A Doubt App ?

Get in Touch

Call / Whats App