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A charge of -2.720 μC is located at (3.000 m , 4.591 m ), and a charge of 1.600 μC is located at (-2.626 m , 0).There is one point on the line connecting these two charges where the potential is zero. Find this point. There is one point on the line connecting these two charges where the potential is zero. Find this point. Express your answers using three decimal places separated by a comma.

x,y = ______ m please help!

Respuesta :

Kazeemsodikisola

Answer:

Explanation:

Given that,

Q1=-2.72μC

At the position r1 =(3 , 4.591)m

Q2=1.600 μC

Located at (-2.626 m , 0)

Let the point where the electric potential will be zero be (x, y)

Therefore,

r1=(3-x, 4.952-y)

r2=(-2.626-x, -y)

Then, electric potential at that point is given as

V1+V2=0

Then,

KQ1/r1 +KQ2/r2=0

Divide through by k

Q1/r1+Q2/r2=0

-2.72/(3-x, 4.952-y) + 1.6/(-2.626-x, -y)=0

2.72/(3-x, 4.952-y) =1.6/(-2.626-x, -y)

For x axis

2.72/(3-x)=1.6/(-2.626-x)

Cross multiply

2.72(-2.626-x)=1.6(3-x)

-7.14272-2.72x=4.8-1.6x

-2.72x+1.6x=4.8+7.14272

-1.12x=11.94272

x=-11.94272/1.12

x=-10.663m

For y axis

2.72/(3-x, 4.952-y) =1.6/(-2.626-x, -y)

2.72/(4.952-y)=1.6/-y

Cross multiply

-2.72y=1.6(4.952-y)

-2.72y=7.9232-1.6y

-2.72y+1.6y=7.9232

-1.12y=7.9232

y=7.9232/-1.12

y=-7.074m

(x, y)=(-10.663, -7.074)m

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