Four charges are phased in a square of ride 2 cm, is shown. What is the potentia

Question

Four charges are phased in a square of ride 2 cm, is shown. What is the potential energy of the system? -4.7J -8.4J -3.6J -9.3J What is the potential at the exact center of the square? 5.1 x 10^6 V 2.5 x 10^6 V 0 V -2.5 x 10^6 V How much work is required to get first point A to point B? -4.7J 0J -4.7J -2.4J If a positive test charge were released at Point A, which direction initially would it move toward? to the right to the left remain motionless toward center of square Five capacitors are connected is series across a potential difference, with the dielectric used, each capacitor will fail if voltage across it exceeds 30.0V. what is highest voltage V_AB that can be supplied? 579 V 69 V 182 V 150 V

Explanation / Answer

a)

There will be 6 interactions:

4 of them will be same and 2 other will be different

So, Utot = - 4*k*q^2/r + 2*k*q^2/(sqrt(2)*r)

= -4*9*10^9*(2*10^-6)^2/0.02 + 2*(9*10^9)*(2*10^-6)^2/(sqrt(2)*0.02)

= -4.65 J

= -4.7 J <-------answer

b)

Potential at center = 0

As the potentials due to 2 of the positive charges are canceled by the potentials due to 2 other negative charges

c)

Potential at point A, Va = 0

Potenial at point B, Vb = 0

So, work done = Va - Vb = 0

d)

They will move towards right as the net force is towards the right side

e)

Charge(Q) across each of the capacitors is same(as they are connected in series)

Now, we know, Q = C*V

So, V = Q/C

So, voltage(C) is inversely proportional to Capacitance(C)

So, maximum volatge will appear across the lowest valued capacitor which is 5uF

Now, Voltage across 5uF , V5 = 30 V

So, voltage across 10uF = V10 = V5/2 = 15 V

voltage across 15uF, V15 = 30*5/15 = 10 V

voltage across 25uF, V25 = 30/5 = 6 V

across 45 uF, V45 = 30/9 = 3.3 V

So, total voltage, Vab = 30+15+10+6+3.3 = 64.3 V = 64 V<-------answer

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