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CBSE Class 12-science Answered

derive an expression for the electric field at a point due to a system  of charges.
Asked by Balan | 26 Mar, 2018, 13:01: PM
answered-by-expert Expert Answer
Let us assume a charge distribution +q1, +q2, -q3, -q4 and q5 which are at distances r1, r2, r3, r4 and r5 respectively from test point P.
 
The electric field at P due to the given charge distribution is the vector sum of electric field due to individual charges.
 
begin mathsize 12px style H e n c e space E space equals space fraction numerator plus q 1 over denominator 4 pi epsilon subscript 0 r subscript 1 superscript 2 end fraction stack space r subscript 1 with hat on top plus fraction numerator plus q 2 over denominator 4 pi epsilon subscript 0 r subscript 2 superscript 2 end fraction space stack space r subscript 2 with hat on top plus fraction numerator negative q 3 over denominator 4 pi epsilon subscript 0 r subscript 3 superscript 2 end fraction space stack space r subscript 3 with hat on top plus fraction numerator negative q 4 over denominator 4 pi epsilon subscript 0 r subscript 4 superscript 2 end fraction space stack space r subscript 4 with hat on top plus fraction numerator plus q 5 over denominator 4 pi epsilon subscript 0 r subscript 5 superscript 2 end fraction stack space r subscript 5 with hat on top h e r e space stack space r subscript k with hat on top space i s space t h e space u n i t space v e c t o r space i n space t h e space d i r e c t i o n space o f space t h e space l i n e space j o i n i n g space k t h space c h a r g e space a n d space t e s t space p o i n t space P I n space g e n e r a l space i f space t h e r e space a r e space n space c h a r g e s space a r e space t h e r e comma space t h e n space E space equals space sum from k equals 1 to n of fraction numerator q subscript k over denominator 4 pi epsilon subscript 0 r subscript k superscript 2 end fraction stack space r subscript k with hat on top end style
Answered by Thiyagarajan K | 27 Mar, 2018, 10:00: AM

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