Given the following bridge: The power supply is set to 3.50 volts. When the left

Question

Given the following bridge: The power supply is set to 3.50 volts. When the left probe end is touched to the 14.6 cm position and the right probe is touching the 46.9 cm position on the bridge: a) What is the electric field strength in the wire, in volts/cm? b) What is the potential difference between the two probes? (c) If the power supply was adjusted to a voltage of 5.00 volts, how would you have to move the left probe from the 14.6 cm position to have the meter read 0.60 volts (that is. leaving the right probe fixed)? (to the left, towards the 0 cm mark, or to the right towards the 100 cm mark) - show why this is so.

Explanation / Answer

a] Magnitude of Electric field strength = V/d

=> E = 3.5/100 = 0.035 Volts/cm

b] Potential difference = - E[14.6 - 46.90] = 1.1305 Volts

c] New Electric field strength = 5/100 = 0.05 Volts/cm

=> Potential difference = 0.6 = - E[x - 46.90]

=> x = 34.90 cm

so the probe must be moved to the right towards the 100 cm mark.

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