Question

four point charges of 10^-7,-10^-7,-2*10-7,2*10^-7 are placed respectively at the corners A,B,C,D of a 0.05m square.find the magnitude of the resultant force on the charge at D.correct and full answer will be marked as brainliest.​

Answer 1

The Magnitude of the resultant force on the charge at D is 25273 x 10⁻²³ N

Explanation:

Given   ;  Side of Square = 0.05 m

             Charge at corner A = $q_{A}$ = 10⁻⁷ C    ;  

              Charge at corner B = $q_{B}$ = -10⁻⁷ C

                 Charge at corner C = $q_{C}$ =     -2 x 10⁻⁷ C

                  Charge at corner D= $q_{D}$ = 2 X 10⁻⁷ C

we have to find magnitude of the resultant force at charge D

According to superposition of forces

so,   $F_{D } = F_{AD} +F_{BD}+F_{CD}$

According to Coulomb's law of forces

$F_{AD}$ = 9 x 10⁻⁹x 10⁻⁷x 2 x 10⁻⁷/(0.05 )²

      = 7200 x 10⁻²³ N

$F_{CD}$  = 9 x 10⁻⁹x (-2 x 10⁻⁷) x 2 x 10⁻⁷ /(0.05)²

        = 14400 x 10⁻²³ N

$F_{BD}$ = BD = $\sqrt{0.05^{2} +0.05^{2} }$ =   0.07 m

= 9 x 10⁻⁹ x (-10⁻⁷) x (2 x 10⁻⁷) / (0.07)²

    = 3673 x 10⁻²³N

Resultant force at charge D = 7200 x 10⁻²³ + 14400 x 10⁻²³+3673 x 10⁻²³

     = 25273 x 10⁻²³ N

Answer 2

Answer:

hope to it will be helpful