Question

question 7 (start on a new page)
three small identical metal spheres, x, y and z, on insulated stands, are initially neutral.
they are then charged to carry charges of -7.5 x 10^-9 c, qy and -3.5 x 10^-9 c respectively,
as shown below.
7.1 define the term electric field strength at a point in words.
7.2 determine by means of a calculation where sphere y must be placed with
reference to sphere x on the left so that it experiences a zero net force.
7.3 how will doubling the charge on sphere x affect the magnitude of the
electrostatic force exerted by spere y on sphere x? give a reason for the
answer.
the charged spheres are brought together so that all three spheres touch each other
at the same time and are then separated. the charge on each sphere, after
separation, is -3 x 10^-9 c.
7.4 determine the value of charge qy.
7.5 were electrons removed from or transferred to y? give a reason
for the answer.
7.6 draw the electric field pattern associated with the charged spheres, y and z,
after they are separated and returned to their original positions.

Explanation:

7.1

Electric field strength at a point is defined as the electrostatic force per unit positive charge experienced at that point.

Step1: Apply Coulomb's law

Let the distance between \(X\) and \(Y\) be \(r\), then the distance between \(Y\) and \(Z\) is \(2 - r\). The force on \(Y\) due to \(X\), \(F_{XY}=k\frac{Q_XQ_Y}{r^{2}}\), and the force on \(Y\) due to \(Z\), \(F_{YZ}=k\frac{Q_YQ_Z}{(2 - r)^{2}}\). Since \(F_{net}=0\), \(F_{XY}=F_{YZ}\), so \(k\frac{Q_XQ_Y}{r^{2}}=k\frac{Q_YQ_Z}{(2 - r)^{2}}\) ( \(Q_Y\) cancels out as \(Q_Y
eq0\)).

Step2: Substitute values

Given \(Q_X=- 7.5\times10^{-9}\space C\), \(Q_Z=-3.5\times10^{-9}\space C\). Then \(\frac{7.5}{r^{2}}=\frac{3.5}{(2 - r)^{2}}\). Cross - multiply: \(7.5(2 - r)^{2}=3.5r^{2}\). Expand \((2 - r)^{2}=4-4r + r^{2}\), so \(7.5(4-4r + r^{2})=3.5r^{2}\). \(30-30r+7.5r^{2}=3.5r^{2}\).

Step3: Simplify the equation

\(7.5r^{2}-3.5r^{2}-30r + 30 = 0\), \(4r^{2}-30r + 30=0\). Using the quadratic formula \(r=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a = 4\), \(b=-30\), \(c = 30\). \(r=\frac{30\pm\sqrt{900-480}}{8}=\frac{30\pm\sqrt{420}}{8}\). \(r=\frac{30\pm20.5}{8}\). We take the positive value \(r = 1.19\space m\) (since \(r<2\)).

According to Coulomb's law \(F = k\frac{Q_1Q_2}{r^{2}}\). If \(Q_1\) (charge on \(X\)) is doubled (\(Q_1'\) = \(2Q_1\)), then \(F'=k\frac{2Q_1Q_2}{r^{2}} = 2F\).

Answer:

Electric field strength at a point is the electrostatic force per unit positive charge at that point.

7.2