Question

Ten positively charged particles are kept fixed on the x axis at points x = 10 cm,20 cm,30 cm,.........,100 cm . The first particle has a charge 1 × ^-8 C,the second one has a charge 8 × 10^-8,the third one 27 × 10^-8 and the tenth particle has a charge 1000 × 10^-8 C. Find the magnitude of the electric force acting on 1 C charge placed at the origin


Thank You​

Answer 1

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To find :

Electirc Force on 1 C charge .

Given data :

1C charge placed at the origin .

at x = 10 cm , first particle, q1 = 1 × ^-8 C

at x = 20 cm , second particle , q2 = 8 ×

^-8 C and x = 100 cm , tenth charge ,q10 = 1000× ^-8 C

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• super postion

Net force on q is given by :

F 1 = F12 + F 13+ F 14 + .....

learn about this , refer to the attachment1 .

• coulombs law ;

${\orange {\boxed{\large{\bold{F =\frac{K q1 q2 }{r{}^{2} }}}}}}$

Fnet = force acting on charge 1 C by the all charges

= F 1 + F2 + F3..... + F 10

${\red{\boxed{\large{\bold{F net \: = 495 × 10{}^{3} N }}}}}$

step by step explanation;

refer to attachment 2&3

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formulas and data used in Question ;

1) 1+2+3+4+5 ... n = n(n+1)/2

2) k= 9× 10^9 Nm²/c²

Answer 2

Answer:

Refer the attachment please..

Hope it helps..

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