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!

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

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .

the answer is 12 as i may remember

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