1840 United States presidential election in Georgia

Main article: 1840 United States presidential election

1840 United States presidential election in Georgia

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Nominee William Henry Harrison Martin Van Buren Party Whig Democratic Home state Ohio New York Running mate John Tyler none Electoral vote 11 0 Popular vote 40,339 31,983 Percentage 55.78% 44.22%

County Results Harrison   50-60%   60-70%   70–80%   80–90%   90–100% Van Buren   50-60%   60-70%   70-80%   80-90%   90-100% Tie   50%

President before election Martin Van Buren Democratic Elected President William Henry Harrison Whig

A presidential election was held in Georgia on November 2, 1840 as part of the 1840 United States presidential election. ^[1] Voters chose 11 representatives, or electors to the Electoral College, who voted for President and Vice President.

Georgia voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Georgia by a margin of 11.56%. This would be the last time that Georgia did not vote for the incumbent Democratic president until 1964.

Results

United States presidential election in Georgia, 1840 [2]
Party		Candidate	Running mate	Popular vote		Electoral vote
				Count	%	Count	%
	Whig	William Henry Harrison of Ohio	John Tyler of Virginia	40,339	55.78%	11	100.00%
	Democratic	Martin Van Buren of New York	Richard M. Johnson of Kentucky	31,983	44.22%	0	0.00%
Total				72,322	100.00%	11	100.00%

References

1. Dubin, Michael J. (2002). United States Presidential Elections, 1788–1860: The Official Results by County and State. Jefferson, NC: McFarland & Co. p. xvi.
2. "1840 Presidential General Election Results - Georgia". U.S. Election Atlas. Retrieved December 23, 2013.