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CBSE Class 12-science Answered

Find the force.
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Asked by pradeepjsme | 27 Jun, 2018, 01:43: PM
answered-by-expert Expert Answer
 
I draw the figure for this question as per my understanding. I understand, there is a cube of side a,  all the corners of cube has a charge Q and the cube corner is placed at origin as shown in my figure. Also I understand that it is required to get force on a test charge q placed on x-axis and at a distance p from origin. In the given figure by the user, this test charge point x-coordinate is shown as a, this can not be,  because the side of cube is already a.
Hence I assumed the x-coordinate of test charge point as p.
 
Force at test charge point is sum of forces due to charges of all 8 corners of cube. In order to understand the distance of each charge from test point, I have given index to all the eight charges, but as per given question all charges are having same magnitude and same sign.
 
The force F on test charge q is written as
 
begin mathsize 12px style F space equals space fraction numerator q over denominator 4 pi epsilon subscript 0 end fraction open curly brackets Q subscript 1 over p squared space plus space fraction numerator Q subscript 2 over denominator p squared plus a squared end fraction plus fraction numerator Q subscript 3 over denominator open parentheses p minus a close parentheses squared plus a squared end fraction space plus space Q subscript 4 over open parentheses p minus a close parentheses squared space plus space fraction numerator Q subscript 5 over denominator open parentheses p minus a close parentheses squared plus a squared end fraction space plus fraction numerator Q subscript 6 over denominator p squared plus a squared end fraction plus space fraction numerator Q subscript 7 over denominator p squared plus 2 a squared end fraction plus fraction numerator Q subscript 5 over denominator open parentheses p minus a close parentheses squared plus 2 a squared end fraction space close curly brackets sin c e space Q subscript 1 equals space Q subscript 2 space equals space Q subscript 3 space equals space Q subscript 4 space equals space Q subscript 5 space equals space Q subscript 6 space equals space Q subscript 7 space equals space Q subscript 8 space equals space Q w e space h a v e comma space space F space equals space fraction numerator q Q over denominator 4 pi epsilon subscript 0 end fraction open curly brackets 1 over p squared space plus space fraction numerator 2 over denominator p squared plus a squared end fraction plus fraction numerator 2 over denominator open parentheses p minus a close parentheses squared plus a squared end fraction space plus space 1 over open parentheses p minus a close parentheses squared space plus space space fraction numerator 1 over denominator p squared plus 2 a squared end fraction plus fraction numerator 1 over denominator open parentheses p minus a close parentheses squared plus 2 a squared end fraction space close curly brackets end style
Answered by Thiyagarajan K | 27 Jun, 2018, 02:45: PM

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