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Cathode ray tubes (CRTs) used in old-style televisions have been replaced by modern LCD and LED screens. Part of the CRT included a set of accelerating plates separated by a distance of about 1.54 cm. If the potential difference across the plates was 26.5 kV, find the magnitude of the electric field (in V/m) in the region between the plates.

Sagot :

akiran007

Answer:

E = 1,720,779.221 or 1.720779221 * 10^ 6V/m

Explanation:

The electric field between the parallel conducting plates is given by

E =V / d

where V is the potential difference and d is the distance between the plates.

E = 26.5 kV/ 1.54 cm

Now we have to convert into proper units

26.5 kv= 26.5 * 1000 v=  26500 volts

1  kv= 1000 volts

1.54 cm = 1.54/ 100 m= 0.0154m

1m = 100cm

Now putting the values

E= 26500/0.0154 = 1,720,779.221 V/m

The Electric field is equal to E= 1,720,799.221 or 1.7220799221 * 10 ^6 Volts per meter.

In scientific notation this can be written as 1.7220799221 *10^6 V/m

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