Dear student The electric field due to a line charge is given by E=14πεo2λr force on charge qF=14πεo2qλrSo force on the negative charge=9×109×2×2×10-8×4×10-42×10-2=7 2 N directed towards the wire force on positive charge=9×109×2×2×10-8×4×10-42 2×10-2=6 5 N directed away from the wirehence net force =7 2-6 5=0 7 N It is directed towards the line charge

Q8

Dear student

T h e e l e c t r i c f i e l d d u e t o a l i n e c h a r g e i s g i v e n b y E = \frac{1}{4 π ε_{o}} \frac{2 λ}{r} f o r c e o n c h a r g e q F = \frac{1}{4 π ε_{o}} \frac{2 q λ}{r} S o f o r c e o n t h e n e g a t i v e c h a r g e = \frac{9 \times 10^{9} \times 2 \times 2 \times 10^{- 8} \times 4 \times 10^{- 4}}{2 \times 10^{- 2}} = 7.2 N d i r e c t e d t o w a r d s t h e w i r e f o r c e o n p o s i t i v e c h a r g e = \frac{9 \times 10^{9} \times 2 \times 2 \times 10^{- 8} \times 4 \times 10^{- 4}}{2.2 \times 10^{- 2}} = 6.5 N d i r e c t e d a w a y f r o m t h e w i r e h e n c e n e t f o r c e = 7.2 - 6.5 = 0.7 N I t i s d i r e c t e d t o w a r d s t h e l i n e c h a r g e .

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