four equal magnitude point charges 3 microcoulomb is placed at the corners of a square that is 40cm inside find the force on any one of the

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four equal magnitude point charges 3 microcoulomb is placed at the corners of a square that is 40cm inside find the force on any one of the charges ​

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MichaelMet 6 months 2021-07-15T22:32:25+00:00 1 Answers 27 views 0

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  1. thaiduong
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    2021-07-15T22:33:39+00:00
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    Answer:

    Approximately 0.97\; \rm N. This force would point away from the center of the square (to the left at 45^\circ above the horizontal direction.)

    Explanation:

    Coulomb’s constant: k \approx 8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2}.

    By Coulomb’s Law, the electrostatic force between two point charges q_1 and q_2 that are separated by r (vacuum) would be:

    \displaystyle F = \frac{k \cdot q_1 \cdot q_2}{r^2}.

    Consider the charge on the top-left corner of this square.

    Apply Coulomb’s Law to find the electrostatic force between this charge and the charge on the lower-left corner.

    Convert quantities to standard units:

    q_1 = q_2 = 3 \times 10^{-6}\; \rm C.

    r = 0.40\; \rm m.

    \begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^2} \\ &\approx \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times (3 \times 10^{-6}\; \rm C)^{2}}{(0.40\; \rm m)^{2}} \\ &\approx 0.506\; \rm N\end{aligned}.

    As the two charges are of the same sign, the electrostatic force on each charge would point away from the other charge. Hence, for the charge on the top-left corner of the square, the electrostatic force from the charge below it would point upwards.

    Similarly, the charge to the right of this charge would exert an electrostatic force with the same magnitude (approximately 0.506\; \rm N) that points leftwards.

    For the charge to the lower-right of the top-left charge, r = \sqrt{2} \times 0.40\; \rm m. Therefore:

    \begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^2} \\ &\approx \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times (3 \times 10^{-6}\; \rm C)^{2}}{(\sqrt{2} \times 0.40\; \rm m)^{2}} \\ &\approx 0.253 \; \rm N\end{aligned}.

    This force would point to the top-left of the top-left charge, which is 45^\circ above the horizontal direction. Decompose this force into two components that are normal to one another:

    • Horizontal component: approximately \sin(45^\circ) \times 0.253\; \rm N \approx 0.179\; \rm N.
    • Vertical component: approximately \cos(45^\circ) \times 0.253\; \rm N \approx 0.179\; \rm N

    Consider the net force on the top-left charge in two components:

    • Horizontal component: approximately 0.506\; \rm N from the charge on the top-right corner and approximately 0.179\; \rm N from the charge on the lower-right corner. Both components point to the left-hand side. F_x \approx 0.506\; \rm N + 0.179\; \rm N = 0.685\;\rm N (to the left).
    • Vertical component: approximately 0.506\; \rm N from the charge on the lower-left corner and approximately 0.179\; \rm N from the charge on the lower-right corner. Both components point upwards. F_y \approx 0.506\; \rm N + 0.179\; \rm N = 0.685\;\rm N (upward).

    Combine these two components to find the magnitude of the net force on this charge:

    \begin{aligned}F &= \sqrt{{F_x}^{2} + {F_y}^{2}} \\ &\approx \sqrt{0.685^2 + 0.685^2 }\; \rm N \\ &\approx 0.97\; \rm N\end{aligned}.

    This force would point to the top-left of this charge (also at 45^\circ above the horizontal direction, away from the center of the square) because its horizontal and vertical components have the same magnitude.

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