91影视

The given data can be listed below as:

• The acceleration due to gravity is\(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).
• The mass of the charge is\(m = 1.0{\rm{ g}}\left( {\frac{{1{\rm{ kg}}}}{{1000{\rm{ g}}}}} \right) = {10^{ - 3}}{\rm{ kg}}\).
• The position of the charge above the lowest position is\(d = 12{\rm{ cm}}\).
• The length of the cord is\(l = 55{\rm{ cm}}\).
• The value of the electric field is\(E = 9500{\rm{ N/C}}\).
• The Coulomb鈥檚 constant value is \(k = 9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}\).

In this problem, the point charge is situated at the end of the cord, and it is inclined at some angle.The weight of the charge acts in the downward direction. The electric force acts in the rightward direction. The tension force acts along the cord.

Resolve the tension force and apply the equilibrium equations to evaluate the magnitude of the charge.

The free body diagram (1) can be shown as:

Here,\(\theta \)is the angle of inclination,\(Q\)is the charge, mg is the weight of the charge, \({F_{\rm{E}}}\) is the electric force, and \({F_{\rm{T}}}\) is the tension force of the cord.

From the given figure 16-66, the vertical length of the cord can be evaluated as:

\(\begin{aligned}{c}x &= l - d\\ &= \left( {55{\rm{ cm}} - {\rm{12 cm}}} \right)\\ &= 43{\rm{ cm}}\end{aligned}\)

The triangle in diagram (2) can be shown as:

Here, length x becomes the perpendicular side of the triangle.

The angle can be evaluated as:

\(\begin{aligned}{c}\cos \theta &= \frac{{AB}}{{AC}}\\\cos \theta &= \frac{{43{\rm{ cm}}}}{{55{\rm{ cm}}}}\\\theta &= {\cos ^{ - 1}}\left( {0.782} \right)\\\theta &= 38.55^\circ \end{aligned}\)

At equilibrium, resolve the forces along the horizontal and vertical axes. The net force acting in the vertical and horizontal directions can be equated to zero.

The force acting along the horizontal direction can be expressed as:

\(\begin{aligned}{c}\sum {{F_x}} &= 0\\{F_{\rm{E}}} - {F_{\rm{T}}}\sin \theta &= 0\end{aligned}\)

\({F_{\rm{E}}} = {F_{\rm{T}}}\sin \theta \) 鈥� (i)

The electric force on the charge can also be expressed as:

\({F_{\rm{E}}} = QE\) 鈥� (ii)

Equate equations (i) and (ii).

\({F_{\rm{T}}}\sin \theta = QE\)

\({F_{\rm{T}}} = \frac{{QE}}{{\sin \theta }}\) 鈥� (iii)

The force acting along the vertical direction can be expressed as:

\(\begin{aligned}{c}\sum {{F_y}} &= 0\\{F_{\rm{T}}}\cos \theta - mg &= 0\end{aligned}\)

\({F_{\rm{T}}} = \frac{{mg}}{{\cos \theta }}\) 鈥� (iv)

Equate equations (iii) and (iv).

\(\begin{aligned}{c}\frac{{QE}}{{\sin \theta }} &= \frac{{mg}}{{\cos \theta }}\\Q &= \frac{{\sin \theta \times mg}}{{\cos \theta \times E}}\\Q &= \frac{{mg\tan \theta }}{E}\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}Q &= \frac{{{{10}^{ - 3}}{\rm{ kg}} \times 9.81{\rm{ m/}}{{\rm{s}}^2} \times \tan 38.55^\circ }}{{9500{\rm{ N/C}}}}\\ &= \frac{{7.82 \times {{10}^{ - 3}}{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2}\left( {\frac{{1{\rm{ N}}}}{{1{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}} \right)}}{{9500{\rm{ N/C}}}}\\ &= 8.23 \times {10^{ - 7}}{\rm{ C}}\end{aligned}\)

The charge should be positive since the electric field and force are in the same direction (rightward).

Thus, the magnitude of the charge is \(8.23 \times {10^{ - 7}}{\rm{ C}}\) and its sign is positive.