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Four particles each having a charge q ar...

Four particles each having a charge q are placed on the four vertices of a regular pentagon. The distance of each comer from !be centre is 'a'. Find the electric field at the centre of pentagon.

Text Solution

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The correct Answer is:
A, B, D
In case of five charges at five vertices of regular pentagon net electric field at centre is zero. Because five vectors of equal magnitudes from a closed regular pentagon as shown in figure

where one charge is removed. Then one vector (let `AB`) is deceased hence the net resultant is equal to magnitude of one vector
`=|BA|=1/(4piepsilon_0) q/a^2`
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Knowledge Check

  • Find each interior of a regular pentagon.

    A
    `108^(@)`
    B
    `105^(@)`
    C
    `103^(@)`
    D
    `101^(@)`

  • Five point charges, +q each are placed at the five vertices of a regular hexagon. The distance of the center of the hexagon from any fo the vertices is a. the electric field at the center of the hexagon is

    A
    `(q)/(4 pi epsilon_(0)a^(2))`
    B
    `(q)/(8 pi epsilon_(0)a^(2))`
    C
    `(q)/(16 pi epsilon_(0)a^(2))`
    D
    zero

  • Five point charge ( +q each) are placed at the five vertices of a regular hexagon of side 2a .what is the magnitude of the net electric field at the centre of the hexazon?

    A
    `1/(4piepsilon_0)q/a^2`
    B
    `q/(16 piepsilon_0a^2)`
    C
    `(sqrt2q)/(4piepsilon_0a^2)`
    D
    `(5q)/(16piepsilon_0a^2)`