Questions: Campo elettrico-2 1. Due cariche puntiformi, q e -4 q, sono separate da una distanza d. In quale posizione il campo elettrico risultante, da esse generato, è nullo? Risolvi ponendo q=1.0 x 10^-6 Ce d=2.0 cm Determina il campo elettrico (modulo, direzione e verso) al centro del rettangolo in figura. Risolvi ponendo q=2.00 x 10^-6 Ce a=1.00 mm

Campo elettrico-2

1. Due cariche puntiformi, q e -4 q, sono separate da una distanza d. In quale posizione il campo elettrico risultante, da esse generato, è nullo? Risolvi ponendo q=1.0 x 10^-6 Ce d=2.0 cm

Determina il campo elettrico (modulo, direzione e verso) al centro del rettangolo in figura. Risolvi ponendo q=2.00 x 10^-6 Ce a=1.00 mm
Transcript text: Campo elettrico-2 1. [3 punti] Due cariche puntiformi, $q$ e $-4 q$, sono separate da una distanza $d$. In quale posizione il campo elettrico risultante, da esse generato, è nullo? Risolvi ponendo $q=1.0 \times 10^{-6} \mathrm{Ce} d=2.0 \mathrm{~cm}$ [5 punti] Determina il campo elettrico (modulo, direzione e verso) al centro del rettangolo in figura. Risolvi ponendo $q=2.00 \times 10^{-6} \mathrm{Ce} a=1.00 \mathrm{~mm}$
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Solution

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Solution Steps

Step 1: Find the null point for the electric field.

Let the charge \(q\) be at the origin (x=0) and the charge \(-4q\) be at \(x=d\). We are looking for a point where the electric field due to both charges is zero. Let this point be at a distance \(x\) from the charge \(q\).

If the point lies between the two charges (\(0 < x < d\)), the electric fields due to both charges will point in the same direction and cannot cancel each other.

If the point lies to the left of the charge \(q\) (\(x < 0\)), let the distance from \(q\) be \(|x|\). The distance from \(-4q\) will be \(d + |x|\). The electric field due to \(q\) is given by \(E_1 = kq/x^2\) and the electric field due to \(-4q\) is given by \(E_2 = -4kq/(d+x)^2\). For the net field to be zero, \(E_1 + E_2 = 0\), which gives:

\(kq/x^2 = 4kq/(d+x)^2\) \(1/x^2 = 4/(d+x)^2\) \((d+x)^2 = 4x^2\) \(d+x = \pm 2x\)

Case 1: \(d+x = 2x\) implies \(x = d\). This is between the charges and not a valid solution. Case 2: \(d+x = -2x\) implies \(3x = -d\) and \(x = -d/3\). This is a valid solution.

If the point lies to the right of \(-4q\) (\(x > d\)), then the distance from \(q\) is \(x\) and from \(-4q\) is \(x-d\). The electric field due to \(q\) is \(E_1 = kq/x^2\) and the electric field due to \(-4q\) is \(E_2 = -4kq/(x-d)^2\). Setting \(E_1 + E_2 = 0\) gives:

\(kq/x^2 = 4kq/(x-d)^2\) \((x-d)^2 = 4x^2\) \(x-d = \pm 2x\)

Case 1: \(x-d = 2x\) implies \(x = -d\). This places the point to the left of \(q\), not to the right of \(-4q\). Case 2: \(x-d = -2x\) implies \(3x = d\), so \(x = d/3\). But \(d/3\) is between the two charges, not to the right of \(-4q\). So, this is not a valid solution.

Therefore, the only valid solution is \(x=-d/3\). Substituting the values \(q=1.0 \times 10^{-6} C\) and \(d=2.0 cm\), we have \(x = -2.0/3 cm = -0.67 cm\).

Step 2: Calculate Electric Field at the center of the rectangle

Let the charges be \(q_1=q\), \(q_2=2q\), \(q_3=3q\), and \(q_4=4q\). The distance from the center to each corner is \(r = \sqrt{(a/2)^2+(a/2)^2} = a/\sqrt{2}\).

Electric field due to \(q_1\) is \(E_1 = kq/(a^2/2)\) towards \(q_1\). Electric field due to \(q_2\) is \(E_2 = 2kq/(a^2/2)\) towards \(q_2\). Electric field due to \(q_3\) is \(E_3 = 3kq/(a^2/2)\) towards \(q_3\). Electric field due to \(q_4\) is \(E_4 = 4kq/(a^2/2)\) towards \(q_4\).

The x-components of \(E_1\) and \(E_3\) cancel out, and the x-components of \(E_2\) and \(E_4\) add up to \(-(6kq/(a^2/2))(1/\sqrt{2}) = -12\sqrt{2}kq/a^2\) along the negative x-axis.

The y-components of \(E_2\) and \(E_4\) cancel out, and the y-components of \(E_1\) and \(E_3\) add up to \(-(4kq/(a^2/2))(1/\sqrt{2}) = -8\sqrt{2}kq/a^2\) along the negative y-axis.

The resultant electric field has a magnitude \(E = \sqrt{E_x^2 + E_y^2} = (2kq/a^2)\sqrt{72+32} = 20kq/\sqrt{2}a^2 = 10\sqrt{2}kq/a^2\) The direction is \(\theta = \arctan(E_y/E_x) = \arctan(8\sqrt{2}/12\sqrt{2}) = \arctan(2/3)\) relative to the negative x-axis in the third quadrant.

Final Answer

  1. \(x = -d/3 = -0.67 \text{ cm}\)
  2. \(E = 10\sqrt{2}kq/a^2\) where \(q=2.00 \times 10^{-6} C\) and \(a=1.00 \text{ mm}\). Direction: \(\arctan(2/3)\) South of West.
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