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Plzzzz someone explain me how to plot the graphs for two particle charges for the variation of electric field v/s x or y coordiante. Plzzz explain carefully with certain examples. Electric field v/s x &y coordinates; Potential v/s x &y coordinates For eg, if two identical positive point charges q are placed on the axis at x=-a and x = +a. Then how can we plot the variation of electric field v/s x & y ccoordinates Plzzzz someone explain me how to plot the graphs for two particle charges for the variation of electric field v/s x or y coordiante. Plzzz explain carefully with certain examples. Electric field v/s x &y coordinates; Potential v/s x &y coordinates For eg, if two identical positive point charges q are placed on the axis at x=-a and x = +a. Then how can we plot the variation of electric field v/s x & y ccoordinates
Plzzzz someone explain me how to plot the graphs for two particle charges for the variation of electric field v/s x or y coordiante. Plzzz explain carefully with certain examples.
Electric field v/s x &y coordinates;
Potential v/s x &y coordinates
For eg, if two identical positive point charges q are placed on the axis at x=-a and x = +a. Then how can we plot the variation of electric field v/s x & y ccoordinates
hi, it's very easy just notice that E is vector but V is scalar. just mark q at both -a & +a , u'll will see (mathematically/physically) that ,at the points nearer to RHS of -a the distance tends to zero hence E tends to ∞( it 's +∞ bcz distace from -a will b + if we r on RHS of a point) at the mid (at 0) E wud b zero as E of both cancels each other and LHS of +a it wud -∞(near to +a chrge at this pt wud dominate over the othr as r tends to zero E due to tis charge wud tends to ∞ & distance from this pt to any point on LHS wud b - ) similarly on LHS of -a r(distace) is -ve &just on LHS of -a E wud tends to-∞ as r tends to 0 & E wud tend to 0 at x= -∞ onRHS of +a first E ->+∞ just on RHS & on going further it wud decrease & -> 0 at x= +∞ the graph wud b continuous LHS of -a, b/w -a &a and RHS of a it wud b discont. at -a &a only . so will b the V graph vs x (u can use the relation dV/Dx= -E(x) )
hi,
it's very easy just notice that E is vector but V is scalar.
just mark q at both -a & +a , u'll will see (mathematically/physically) that ,at the points nearer to RHS of -a the distance tends to zero hence E tends to ∞( it 's +∞ bcz distace from -a will b + if we r on RHS of a point)
at the mid (at 0) E wud b zero as E of both cancels each other
and LHS of +a it wud -∞(near to +a chrge at this pt wud dominate over the othr as r tends to zero E due to tis charge wud tends to ∞ & distance from this pt to any point on LHS wud b - )
similarly on LHS of -a
r(distace) is -ve &just on LHS of -a E wud tends to-∞ as r tends to 0 & E wud tend to 0 at x= -∞
onRHS of +a
first E ->+∞ just on RHS & on going further it wud decrease & -> 0 at x= +∞
the graph wud b continuous LHS of -a, b/w -a &a and RHS of a
it wud b discont. at -a &a only .
so will b the V graph vs x (u can use the relation dV/Dx= -E(x) )
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points won -