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A square of side length {sidecm} cm has charges of different magnitudes placed at its four corners. Clockwise from the top right as viewed from above these are $q_1$ to $q_4$, with values {q1}, {q2}, {q3} and {q4} nC. In this question, you are required to determine the total electric field, including both magnitude and direction, at the centre of the square due to this arrangement of point charges.

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This question is a relatively complicated example of application of the principle of superposition.

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You should use the values of the physical constants provided in the module notes, assume the medium is air, quote all answers to 4 significant figures and you may express values in scientific format.

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This is a problem of superposition of four different fields, which amounts to the sum of four vectors.  Each field is an expression of Coulomb's Law, so that the pre-factor is

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$\\displaystyle {q_i\\over 4\\pi\\varepsilon_0 r^2}$,

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where since the point of interest is the centre of the square, the distance $r$ is the same for each point-charge (in this case {side*100}/$\\sqrt{2}\\approx$ {siground(r*100,4)} cm). In each case the field points radially away from the point-charge so that for $q_1$ it is down and left in equal measures, for $q_2$ it is up and to the left, and so on.  For negative charges the sign of $q$ can be thought of reversing the direction (fields converge on negative point charges).  Take care in determining the direction and note that diagrams generally help!

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The value of $E$ can be readily obtained and the direction in each case given by

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$\\displaystyle{1\\over\\sqrt{2}}\\left(\\pm \\hat{i} \\pm \\hat{j}\\right)$, 

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with the combinations of $\\pm$ depending upon which charge we're considering and the $\\sqrt{2}$ factor coming from the fact that this is a unit vector.

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Once each vector field has been obtained they can be summed by component (apply the principle of superposition) giving the total $E$-field at the centre of the square.

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Finally the angle with the line joining $q_2$ and $q_4$ can be obained by noting that the dot product of the total field with the direction of the displacement from $q_2$ to $q_4$ yields an angle:

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$\\displaystyle{\\vec{E}.\\left(-\\frac{\\hat{i}}{\\sqrt{2}}+\\frac{\\hat{j}}{\\sqrt{2}}\\right)=\\left|\\vec{E}\\right|\\cos(\\theta)}$

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so

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$\\displaystyle{-E_x+E_y=\\sqrt{2}\\left|\\vec{E}\\right|\\cos(\\theta)\\Rightarrow \\theta=\\cos^{-1}\\left(\\frac{E_y-E_x}{\\sqrt{2}\\left|\\vec{E}\\right|}\\right)}$.

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It then only remains to determine the value of theta that lies in the required angle range.

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$y$-component of the E-field from q1(N/C).

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Point charge in the bottom left corner, nC.

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Relative permittivity of air.

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Magnitude of the field at the centre.

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Length of side of square in cm.

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$x$-component of the E-field from q2 (N/C).

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Angle between the line between q2 and q4 and the E-field

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$y$-component of the E-field from q4 (N/C).

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$y$-component of the E-field due to q2 (N/C).

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$x$-component of the E-field from q4 (N/C).

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Point charge in the top left corner, nC.

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Distance from the charges to the centre of the square in m.

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Point charge in bottom right corner, nC.

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$x$-component of the E-field from q1 (N/C).

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Permittivity of free space in F/m to 4 significant figures.

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The constant, $1/4\\pi\\varepsilon$ in units of m/F.

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$y$-component of the E-field from q3 (N/C).

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$x$-component of the E-field from q3 (N/C).

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Point charge in top right corner in nC.

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r^2, cm

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Side length in m.

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q1 in C

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q2 in C

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q3 in C

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q4 in C

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Determine the electric field at the centre of the square due to $q_1$ ({q1}nC).

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$E_x(q_1)=$[[0]]V/m

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$E_y(q_1)=$[[1]]V/m

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Determine the electric field at the centre of the square due to $q_2$ ({q2}nC).

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$E_x(q_2)=$[[0]]V/m

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$E_y(q_2)=$[[1]]V/m

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Determine the electric field at the centre of the square due to $q_3$ ({q3}nC).

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$E_x(q_3)=$[[0]]V/m

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$E_y(q_3)=$[[1]]V/m

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Determine the electric field at the centre of the square due to $q_4$ ({q4}nC).

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$E_x(q_4)=$[[0]]V/m

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$E_y(q_4)=$[[1]]V/m

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Determine the total electric field at the centre of the square due to the all four charges.

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$E_x=$[[0]]V/m

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$E_y=$[[1]]V/m

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What is the angle, $\\theta$, between the direction of the $E$-field and the line joining the charges $q_2$ ({q2}nC) and $q_4$ ({q4}nC)?  Express your answer as an angle in degrees in the range $0^\\circ\\le\\theta\\le90^\\circ$.

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$\\theta=$[[0]]$^\\circ$.

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