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If 3 different but eqeul charges are forming an equilateral triangle, then where inside the triangle the potential would be 0 and why????? Pls explain!!!!!!! If 3 different but eqeul charges are forming an equilateral triangle, then where inside the triangle the potential would be 0 and why????? Pls explain!!!!!!!
hi, potential due to three chrges at a point is= -kq/r1-kq/r2-kq/r3. thus potential=-kq(1/r1+1/r2+1/r3) if potential has to be zero then that point should be at infinity ( r1r2r3 should be at infinity) plzz approve if you like the answer
hi,
potential due to three chrges at a point is= -kq/r1-kq/r2-kq/r3.
thus potential=-kq(1/r1+1/r2+1/r3)
if potential has to be zero then that point should be at infinity ( r1r2r3 should be at infinity)
plzz approve if you like the answer
dear shiven , first u draw the figure according to these steps 1)draw a equilateral triangle ABC such that its base AB is on x axis , center of base is at origin...third vertice (C) is on y axis ... 2)y axis divides the base in two halves , AO ,BO ... 3) consider any point P(0,y) on y axis then 4) place -q , -q at A,B & q at C... now , potential at P will be V = k(-q)/AP + K-(q)/BP + kq/PC .......1 AO = BO = a/2 PC = asin60 - y = [aroot3/2 - y] ........2 AP = BP = [y^2 + a^2/4]^1/2 ......3 ( by pythagorus theorem) for potential to become 0 , eq 1 = 0 so q/AP + q/BP = q/PC 2/AP = 1/PC AP = 2pC .........4 putting values of AP,PC from eq 2 & 3 we get a quadratic expression 3y^2 - (4root3a)y + 11a^2/4 = 0 from here Y = a[4root3 +(-) root15]/6 = a[ 4root3 - root15]/6 or [4root3 + root15]/6 ( not acceptable coz this point lies ouside) therefore the only point is Y = a[4root3-root15]/6 this is the required point where potential will be zero ...
dear shiven , first u draw the figure according to these steps
1)draw a equilateral triangle ABC such that its base AB is on x axis , center of base is at origin...third vertice (C) is on y axis ...
2)y axis divides the base in two halves , AO ,BO ...
3) consider any point P(0,y) on y axis then
4) place -q , -q at A,B & q at C...
now , potential at P will be
V = k(-q)/AP + K-(q)/BP + kq/PC .......1
AO = BO = a/2
PC = asin60 - y = [aroot3/2 - y] ........2
AP = BP = [y^2 + a^2/4]^1/2 ......3 ( by pythagorus theorem)
for potential to become 0 , eq 1 = 0 so
q/AP + q/BP = q/PC
2/AP = 1/PC
AP = 2pC .........4
putting values of AP,PC from eq 2 & 3 we get a quadratic expression
3y^2 - (4root3a)y + 11a^2/4 = 0
from here Y = a[4root3 +(-) root15]/6
= a[ 4root3 - root15]/6 or [4root3 + root15]/6 ( not acceptable coz this point lies ouside)
therefore the only point is Y = a[4root3-root15]/6
this is the required point where potential will be zero ...
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