For given figure direction of electric field at A will be

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1. Electric Charges and Fields

medium

For the given figure the direction of electric field at $A$ will be

towards $AL$

towards $AY$

towards $AX$

towards $AZ$

Solution

Electric field at $A$ due to $+Q$ at $B$ is given by, $E_{A B}$ in the direction of $X .$

Also, electric field at $C$ is given by $E_{A C}$ in the direction of $Z$.

Hence, the resultant will be the sum of these two electric field and will be given by the parallelogram law of vector addition and will be in the direction of $A Y$.

Standard 12

Physics

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The electric field intensity at a point in vacuum is equal to

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A thin disc of radius $b = 2a$ has a concentric hole of radius $'a'$ in it (see figure). It carries uniform surface charge $'\sigma '$ on it. If the electric field on its axis at height $'h'$ $(h << a)$ from its centre is given as $'Ch'$ then value of $'C'$ is

hard

(JEE MAIN-2015)

Five point charge each having magnitude $‘q’$ are placed at the corner of hexagon as shown in fig. Net electric field at the centre $‘O’$ is $\vec E$. To get net electric field at $‘O’$ be $6\vec E$, charge placed on the remaining sixth corner should be

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A ring of radius $R$ is charged uniformly with a charge $+\,Q$ . The electric field at a point on its axis at a distance $r$ from any point on the ring will be

medium

For the given figure the direction of electric field at $A$ will be

Solution

Similar Questions

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