Two point charges $$q110$$μc and $$q2$$−$$25$$μc are placed on the x−axis at $$x=1m$$ and $$x=4m$$ respectively the electric field in $$V/m$$ at point $$y=3m$$ on y−axis is $$take14$$π∈$$0=9$$×$$109$$ N−$$m2/C2$$

Question

Two point charges $$q110$$μc  and  $$q2$$−$$25$$μc are placed on the  x−axis at $$x=1m$$  and  $$x=4m$$ respectively the electric field in $$V/m$$  at point $$y=3m$$ on y−axis  is  $$take14$$π∈$$0=9$$×$$109$$ N−$$m2/C2$$

Infinity Learn · Accepted Answer

Let   $$E1$$ ​ &   $$E2$$ ​ are the values of electric field due to $$q1$$ ​and $$q2$$ ​ respectively  Magnitude of $$E2$$ $$=q24$$π∈$$r2$$ ⇒$$E2=$$ $$9$$×$$109$$×$$25$$×$$10-6(42+32)$$ ⇒$$E2$$ $$=$$ $$9$$×$$103$$ In vector form, $$E2$$ $$=$$ $$9$$×$$103$$ $$($$$$cos$$θ$$2$$ $$i^$$ $$-$$ $$sin$$θ$$2$$ $$j^)$$  using trigonometry, $$cos$$θ$$2$$ $$=$$ $$4/5$$  and  $$sin$$θ$$2$$ $$=$$ $$3/5$$ Hence, $$E2$$ $$=$$ (72i^-54j^) ×$$102$$  Magnitude of $$E1$$ $$=q14$$π∈$$r2$$ ⇒$$E1=$$ $$9$$×$$109$$×$$10$$×$$10-6(12+32)$$  ⇒$$E1$$ $$=$$ $$910$$×$$102$$  In vector form, $$E1$$ $$=$$ $$910$$×$$102$$ $$(-$$$$cos$$θ$$1$$ $$i^$$ $$+$$ $$sin$$θ$$1$$ $$j^)$$  using trigonometry, $$cos$$θ$$1$$ $$=$$ $$1/10$$  and  $$sin$$θ$$1$$ $$=$$ $$3/10$$ Hence, $$E1$$ $$=$$ $$(-9i^+27j^)$$ ×$$102$$  So, net electric field is the vector sum $$Enet=E1+E2$$         $$=$$ (63i^-27j^) ×$$102$$

Two point charges q110μc and q2−25μc are placed on the x−axis at x=1m and x=4m respectively the electric field in V/m at point y=3m on y−axis is take14π∈0=9×109 N−m2/C2

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